{"id":840,"date":"2019-07-23T20:15:39","date_gmt":"2019-07-23T12:15:39","guid":{"rendered":"https:\/\/xbtzone.com\/blog\/?p=840"},"modified":"2020-05-20T11:49:54","modified_gmt":"2020-05-20T03:49:54","slug":"day-64-%e6%a0%91%e5%bd%a2%e6%95%b0%e6%8d%ae%e7%bb%93%e6%9e%84%e4%b9%8b%e7%ba%bf%e6%ae%b5%e6%a0%91","status":"publish","type":"post","link":"https:\/\/xbtzone.com\/blog\/day-64-%e6%a0%91%e5%bd%a2%e6%95%b0%e6%8d%ae%e7%bb%93%e6%9e%84%e4%b9%8b%e7%ba%bf%e6%ae%b5%e6%a0%91\/","title":{"rendered":"Day 64 \u6811\u5f62\u6570\u636e\u7ed3\u6784\u4e4b\u7ebf\u6bb5\u6811"},"content":{"rendered":"\n

\u7ebf\u6bb5\u6811\u662f\u4e00\u79cd\u57fa\u672c\u7684\u9ad8\u7ea7\u6570\u636e\u7ed3\u6784\uff0c\u5b66\u4e60\u5b83\u5bf9\u89e3\u51b3\u533a\u95f4\u95ee\u9898\u6709\u5f88\u5927\u5e2e\u52a9<\/p>\u8fd9\u7bc7\u6587\u7ae0\u662f\u6211\u5bf9\u4e0a\u5b66\u671fACM-ICPC\u9009\u4fee\u8bfe\u7a0b\u7ed3\u8bfe\u8bba\u6587\u7684\u5c0f\u6574\u7406\u3002<\/cite><\/blockquote>\n\n\n\n

\n\n\n\n

1 \u6982\u8ff0<\/h2>\n\n\n\n

\u7ebf\u6bb5\u6811\uff08Segment tree\uff09\u53c8\u540d\u533a\u95f4\u6811\uff0c\u662f\u4e00\u79cd\u9ad8\u7ea7\u6570\u636e\u7ed3\u6784\uff0c\u5b83\u57fa\u4e8e\u4e8c\u53c9\u641c\u7d22\u6811\u3002\u7ebf\u6bb5\u6811\u7ecf\u5e38\u7528\u4e8e\u89e3\u51b3\u533a\u95f4\u95ee\u9898\u548c\u7ebf\u6bb5\u95ee\u9898\uff0c\u5b83\u7684\u4f18\u70b9\u5728\u4e8e\u80fd\u5feb\u901f\u5730\u67e5\u8be2\u6216\u4fee\u6539\u67d0\u533a\u95f4\u8303\u56f4\u7684\u6307\u5b9a\u6807\u8bb0\u503c\u3002<\/p>\n\n\n\n

\u5b9e\u73b0\u7ebf\u6bb5\u6811\u8fd9\u4e00\u6570\u636e\u7ed3\u6784\u7684\u57fa\u672c\u601d\u60f3\uff1a\u4e8c\u5206\u6cd5\u548c\u9012\u5f52<\/strong>\u3002<\/p>\n\n\n\n

\u7ebf\u6bb5\u6811\u57fa\u4e8e\u4e8c\u53c9\u6811\u5f62\u7ed3\u6784\uff0c\u4e00\u822c\u91c7\u7528\u6570\u7ec4\u7ed3\u6784\u50a8\u5b58\uff0c\u6811\u7684\u6bcf\u4e2a\u7ed3\u70b9\u5305\u542b\u7684\u4fe1\u606f\u6709\uff1a<\/p>\n\n\n\n

\uff081\uff09\u533a\u95f4\u5de6\u7aef\u70b9<\/p>\n\n\n\n

\uff082\uff09\u533a\u95f4\u53f3\u7aef\u70b9<\/p>\n\n\n\n

\uff083\uff09\u6807\u8bb0\u7684\u6570\u636e<\/p>\n\n\n\n

\uff084\uff09\uff08\u53ef\u9009\uff09\u61d2\u6807\u8bb0\u3002<\/p>\n\n\n\n

\u4e00\u68f5\u8868\u793a\u533a\u95f4[1,4]\u7684\u7ebf\u6bb5\u6811\u53ef\u5982\u56fe1\u6240\u793a\u3002<\/p>\n\n\n\n

\"\"
\u56fe1<\/figcaption><\/figure><\/div>\n\n\n\n

\u7ebf\u6bb5\u6811\u5177\u6709\u4ee5\u4e0b\u6027\u8d28\uff1a<\/p>\n\n\n\n

\u6027\u8d28<\/strong> 1 <\/strong> \u8bbe\u7ed3\u70b9\u6240\u8868\u793a\u7684\u533a\u95f4\u8303\u56f4\u4e3a[left<\/em>,right<\/em>],\u5219\u4e2d\u70b9\u5750\u6807mid<\/em> = (left<\/em> + right<\/em>) \u00f7 2\uff0c\u5b83\u7684\u5de6\u5b50\u7ed3\u70b9\u6240\u8868\u793a\u7684\u533a\u95f4\u8303\u56f4\u4e3a[left<\/em>,mid<\/em>]\uff0c\u53f3\u5b50\u7ed3\u70b9\u6240\u8868\u793a\u7684\u533a\u95f4\u8303\u56f4\u4e3a[mid<\/em> + 1,right<\/em>]\u3002<\/p>\n\n\n\n

\u6027\u8d28 2 <\/strong> \u82e5\u6709\u4e00\u7ed3\u70b9k<\/em>\uff0ck<\/em>\u4e3a\u5176\u5728\u6570\u7ec4\u4e2d\u7684\u4f4d\u7f6e\uff0c\u5176\u5de6\u5b50\u7ed3\u70b9\u5728\u6570\u7ec4\u4e2d\u7684\u4f4d\u7f6e\u4e3a2k<\/em>\uff0c\u5176\u53f3\u5b50\u7ed3\u70b9\u5728\u6570\u7ec4\u4e2d\u7684\u4f4d\u7f6e\u4e3a2k<\/em>+1\u3002<\/p>\n\n\n\n

\n\n\n\n

2 \u7ebf\u6bb5\u6811\u7684\u57fa\u672c\u64cd\u4f5c<\/h2>\n\n\n\n

\u4e00\u3001\u5f15\u4f8b<\/strong><\/h3>\n\n\n\n
\"\"<\/figure><\/div>\n\n\n\n

\u5206\u6790<\/strong> <\/strong>\u53ef\u4ee5\u91c7\u7528\u679a\u4e3e\u6cd5\uff0c\u5bf9\u533a\u95f4\u91cc\u9762\u7684\u6bcf\u4e00\u4e2a\u6570\u90fd\u8fdb\u884c\u52a0m<\/em>\u64cd\u4f5c\uff0c\u7b97\u6cd5\u590d\u6742\u5ea6\u4e3aO<\/em>(n<\/em>)\u3002\u4f46\u662fn<\/em>\u7684\u6570\u636e\u8303\u56f4\u592a\u5927\uff0c\u91c7\u7528\u679a\u4e3e\u6cd5\u4f1a\u8d85\u65f6\u3002\u53ef\u4ee5\u91c7\u7528\u7ebf\u6bb5\u6811\u6cd5\u89e3\u51b3\u8be5\u95ee\u9898\u3002<\/p>\n\n\n\n

\u4e8c\u3001\u61d2\u6807\u8bb0<\/h3>\n\n\n\n

\u89e3\u51b3\u4e0a\u8ff0\u95ee\u9898\uff0c\u5982\u679c\u5148\u5bf9\u533a\u95f4[s<\/em>,t<\/em>]\u5185\u7684\u6240\u6709\u53f6\u5b50\u7ed3\u70b9\u6240\u50a8\u5b58\u7684\u6574\u6570\u90fd\u8fdb\u884c\u4e00\u6b21\u52a0m<\/em>\u64cd\u4f5c\uff0c\u518d\u5bf9\u5176\u975e\u53f6\u7236\u7ed3\u70b9\u56de\u6eaf\u6c42\u548c\uff0c\u6700\u540e\u5bf9\u5728\u533a\u95f4\u5185\u7684\u6240\u6709\u6570\u8fdb\u884c\u6c42\u548c\u5e76\u67e5\u8be2\uff0c\u5176\u7b97\u6cd5\u7684\u65f6\u95f4\u590d\u6742\u5ea6\u4f3c\u4e4e\u8981\u6bd4O<\/em>(n<\/em>)\u8fd8\u8981\u9ad8\u5f97\u591a\u3002\u4f46\u6b63\u5982\u6982\u8ff0\u91cc\u9762\u6240\u8bf4\uff0c\u7ebf\u6bb5\u6811\u5e76\u4e0d\u9700\u8981\u5bf9\u67d0\u4e00\u533a\u95f4\u5185\u7684\u6240\u6709\u53f6\u5b50\u7ed3\u70b9\u8fdb\u884c\u4fee\u6539\u548c\u67e5\u8be2\u7684\u64cd\u4f5c\u3002\u5bf9\u4e8e\u5f15\u4f8b\u4e2d\u7684\u95ee\u9898\uff0c\u5728\u4fee\u6539\u533a\u95f4\u6570\u636e\u65f6\uff0c\u53ea\u9700\u8981\u5bf9\u67d0\u4e00\u533a\u95f4\u6240\u5728\u7684\u7ed3\u70b9\u52a0\u4e0a(m<\/em>\u00d7\u533a\u95f4\u957f\u5ea6)\u5373\u53ef\uff0c\u8be5\u533a\u95f4\u7684\u6574\u6570\u548c\u6570\u636e\u5c31\u53ef\u4ee5\u76f4\u63a5\u67e5\u8be2\u51fa\u6765\u4e86\u3002\u4f46\u662f\uff0c\u4fee\u6539\u533a\u95f4\u548c\u67e5\u8be2\u533a\u95f4\u662f\u53ef\u4ee5\u4e0d\u540c\u7684\uff0c\u5728\u4fee\u6539\u533a\u95f4\u6570\u636e\u7ed3\u675f\u540e\u8981\u67e5\u8be2\u533a\u95f4\u6570\u636e\u65f6\uff0c\u5982\u679c\u8981\u67e5\u8be2\u7684\u533a\u95f4\u662f\u4e0a\u8ff0\u4fee\u6539\u533a\u95f4\u7684\u4e00\u4e2a\u5b50\u96c6\uff0c\u8fd9\u65f6\u5c31\u8981\u5bf9\u4e0a\u8ff0\u52a0\u6cd5\u64cd\u4f5c\u4e0b\u53d1\u5230\u5b50\u7ed3\u70b9\uff0c\u5373\u8ba9\u5b50\u7ed3\u70b9\u4e5f\u52a0\u4e0a(m<\/em>\u00d7\u533a\u95f4\u957f\u5ea6)\uff0c\u4e3a\u5b9e\u73b0\u4e0b\u53d1\u64cd\u4f5c\uff0c\u6211\u4eec\u8981\u5f15\u5165\u4e00\u4e2a\u65b0\u7684\u7ed3\u70b9\u4fe1\u606f\uff1a\u61d2\u6807\u8bb0<\/strong>\uff0c\u7528\u4e8e\u7d2f\u79ef\u533a\u95f4\u7ed3\u70b9\u7684\u64cd\u4f5c\u3002\u5728\u67e5\u8be2\u88ab\u4fee\u6539\u6570\u636e\u7684\u533a\u95f4\u7ed3\u70b9\u7684\u5b50\u533a\u95f4\u7ed3\u70b9\u6570\u636e\u65f6\uff0c\u61d2\u6807\u8bb0\u4f1a\u53d1\u6325\u5176\u4e0b\u53d1\u7d2f\u8ba1\u7684\u52a0\u6cd5\u64cd\u4f5c\u529f\u80fd\u3002<\/p>\n\n\n\n

\u540c\u6837\u5730\uff0c\u5982\u679c\u8981\u4fee\u6539\u7684\u533a\u95f4\u662f\u67d0\u4e00\u533a\u95f4\u7684\u5b50\u96c6\uff0c\u8fd9\u65f6\u8981\u628a\u7ed3\u679c\u56de\u6eaf\u5230\u7236\u7ed3\u70b9\uff0c\u4f46\u6b64\u65f6\u5e76\u4e0d\u9700\u8981\u501f\u52a9\u61d2\u6807\u8bb0\uff0c\u56e0\u4e3a\u5728\u67d0\u6b21\u9012\u5f52\u64cd\u4f5c\u7ed3\u675f\u524d\u53ef\u4ee5\u76f4\u63a5\u628a\u7236\u7ed3\u70b9\u6570\u636e\u8d4b\u503c\u4e3a\u5de6\u53f3\u5b50\u7ed3\u70b9\u7684\u6570\u636e\u4e4b\u548c\u3002<\/p>\n\n\n\n

\u4e09\u3001\u5efa\u7acb\u7ebf\u6bb5\u6811<\/h3>\n\n\n\n

\u7ebf\u6bb5\u6811\u57fa\u4e8e\u4e8c\u53c9\u641c\u7d22\u6811\uff0c\u5176\u6570\u636e\u7ed3\u6784\u4e0e\u4e8c\u53c9\u6811\u5f62\u6570\u636e\u7ed3\u6784\u7c7b\u4f3c\u3002<\/p>\n\n\n\n

\u672c\u4f8b\u91c7\u7528\u6570\u7ec4\u6765\u50a8\u5b58\u7ebf\u6bb5\u6811\u7684\u7ed3\u70b9\uff0c\u6570\u7ec4\u957f\u5ea6\u4e3a4n<\/em>+1\u3002<\/p>\n\n\n\n

#define max_n 1000000<\/p>

typedef struct _tNode<\/p>

{<\/p>

    int l;<\/p>

    int r;<\/p>

    int lazy;<\/p>

    int sum;<\/p>

}Node;<\/p>

Node node[4 * max_n + 1];<\/p><\/blockquote>\n\n\n\n

\u7ed3\u5408\u6027\u8d281\u548c\u6027\u8d282\uff0c\u5229\u7528\u4e8c\u5206\u6cd5\u548c\u9012\u5f52\u601d\u60f3\uff0c\u53ef\u4ee5\u5f88\u5bb9\u6613\u5730\u5efa\u7acb\u4e00\u68f5\u7ebf\u6bb5\u6811\u3002k<\/em>\u8868\u793a\u6811\u6839\u7684\u6570\u7ec4\u4e0b\u6807\uff0cl<\/em>\u3001r<\/em>\u5206\u522b\u8868\u793a\u533a\u95f4\u7684\u5de6\u53f3\u7aef\u70b9\u3002<\/p>\n\n\n\n

void build(int k,int l,int r)<\/p>

{<\/p>

    if (l > r) <\/p>

       return;<\/p>

    node[k].l = l;<\/p>

    node[k].r = r;<\/p>

    node[k].lazy = 0;<\/p>

    if (l == r)<\/p>

    {<\/p>

       node[k].sum = 1;<\/p>

       return;<\/p>

    }<\/p>

    int mid = (l + r) \/ 2;<\/p>

    build(k * 2, l, mid);<\/p>

    build(k * 2 + 1, mid + 1, r);<\/p>

    up(k);<\/p>

}<\/p><\/blockquote>\n\n\n\n

up\u51fd\u6570\u6267\u884c\u5de6\u53f3\u5b50\u7ed3\u70b9\u6570\u636e\u4e4b\u548c\u56de\u6eaf\u7236\u7ed3\u70b9\u64cd\u4f5c\uff0c\u5176\u4ee3\u7801\u5982\u4e0b\u3002<\/p>\n\n\n\n

void up(int k)<\/p>

{<\/p>

    node[k].sum = node[k * 2].sum + node[k * 2 + 1].sum;<\/p>

}<\/p><\/blockquote>\n\n\n\n

\u56db\u3001\u7ebf\u6bb5\u6811\u533a\u95f4\u67e5\u8be2<\/h3>\n\n\n\n

\u4ece\u6839\u7ed3\u70b9\u5f00\u59cb\uff0c\u5224\u65ad\u7ed3\u70b9\u6240\u8868\u793a\u7684\u533a\u95f4\u8303\u56f4\u662f\u5426\u5728\u6240\u8981\u67e5\u8be2\u7684\u533a\u95f4\u8303\u56f4\u4ee5\u5185\uff0c\u5982\u679c\u662f\u5219\u8fd4\u56de\u7ed3\u70b9\u6240\u8868\u793a\u7684\u533a\u95f4\u7684\u548c\u3002\u5426\u5219\uff0c\u5c06\u61d2\u6807\u8bb0\u4ece\u7236\u7ed3\u70b9\u4e0b\u53d1\u81f3\u5de6\u53f3\u5b50\u7ed3\u70b9\u3002\u88ab\u67e5\u8be2\u533a\u95f4\u6240\u5728\u7684\u7ed3\u70b9\u53ef\u80fd\u6a2a\u8de8\u4e24\u4e2a\u6216\u4e24\u4e2a\u4ee5\u4e0a\u7684\u7ed3\u70b9\uff0c\u56e0\u6b64\u5728\u4e0b\u4e00\u6b65\uff0c\u5982\u679c\u67e5\u8be2\u533a\u95f4\u7684\u5de6\u7aef\u70b9\u5c0f\u4e8e\u6216\u7b49\u4e8e\u5f53\u524d\u7ed3\u70b9\u533a\u95f4\u7684\u4e2d\u70b9\u65f6\uff0c\u9012\u5f52\u7ed3\u70b9\u7684\u5de6\u5b50\u6811\uff0c\u5982\u679c\u67e5\u8be2\u533a\u95f4\u7684\u53f3\u7aef\u70b9\u5927\u4e8e\u5f53\u524d\u7ed3\u70b9\u533a\u95f4\u7684\u4e2d\u70b9\u65f6\uff0c\u9012\u5f52\u7ed3\u70b9\u7684\u53f3\u5b50\u6811\u3002\u6700\u540e\u5c06\u5f97\u5230\u7684\u5de6\u5b50\u6811\u7684\u6574\u6570\u548c\u4e0e\u53f3\u5b50\u6811\u7684\u6574\u6570\u548c\u76f8\u52a0\u5e76\u8fd4\u56de\uff0c\u5373\u53ef\u5f97\u5230\u5f85\u67e5\u533a\u95f4\u7684\u548c\u3002<\/p>\n\n\n\n

int query(int k,int l,int r)<\/p>

{<\/p>

    int sum_l = 0, sum_r = 0;<\/p>

    if (node[k].l >= l && node[k].r <= r)<\/p>

       return node[k].sum;<\/p>

    down(k);<\/p>

    int mid = (node[k].l + node[k].r) \/ 2;<\/p>

    if (l <= mid)<\/p>

       sum_l = query(k * 2, l, r);<\/p>

    if (r > mid)<\/p>

       sum_r = query(k * 2 + 1, l, r);<\/p>

    return sum_l + sum_r;<\/p>

}<\/p><\/blockquote>\n\n\n\n

\u7236\u7ed3\u70b9\u4e0b\u53d1\u61d2\u6807\u8bb0\u5230\u5de6\u53f3\u5b50\u7ed3\u70b9\u65f6\uff0c\u5982\u679c\u7236\u7ed3\u70b9\u65e0\u61d2\u6807\u8bb0\u8bb0\u5f55\uff0c\u5219\u76f4\u63a5\u8fd4\u56de\uff0c\u5982\u679c\u6709\uff0c\u5219\u628a\u5176\u5de6\u53f3\u5b50\u7ed3\u70b9\u7684\u61d2\u6807\u8bb0\u8d4b\u503c\u4e3a\u7236\u7ed3\u70b9\u7684\u61d2\u6807\u8bb0\uff0c\u518d\u5c06\u5de6\u53f3\u5b50\u7ed3\u70b9\u6240\u5b58\u50a8\u7684\u6570\u8d4b\u503c\u4e3a\u61d2\u6807\u8bb0\u4e0e\u5de6\u53f3\u5b50\u7ed3\u70b9\u5404\u81ea\u6240\u4ee3\u8868\u7684\u533a\u95f4\u957f\u5ea6\u4e4b\u79ef\uff0c\u6700\u540e\u5c06\u7236\u7ed3\u70b9\u7684\u61d2\u6807\u8bb0\u6e05\u7a7a\u3002<\/p>\n\n\n\n

void down(int k)<\/p>

{<\/p>

    if (node[k].lazy)<\/p>

    {<\/p>

       node[k * 2].lazy += node[k].lazy;<\/p>

       node[k * 2 + 1].lazy += node[k].lazy;<\/p>

       node[k * 2].sum += node[k].lazy * (node[k * 2].r – node[k * 2].l + 1);<\/p>

       node[k * 2 + 1].sum += node[k].lazy * (node[k * 2 + 1].r – node[k * 2 + 1].l + 1);<\/p>

       node[k].lazy = 0;<\/p>

    }<\/p>

}<\/p><\/blockquote>\n\n\n\n

\u4e94\u3001\u7ebf\u6bb5\u6811\u533a\u95f4\u4fee\u6539<\/h3>\n\n\n\n

\u4e0e\u7ebf\u6bb5\u6811\u7684\u533a\u95f4\u67e5\u8be2\u7c7b\u4f3c\uff0c\u4ece\u6839\u7ed3\u70b9\u5f00\u59cb\uff0c\u5982\u679c\u5f53\u524d\u7ed3\u70b9\u6240\u8868\u793a\u7684\u533a\u95f4\u5728\u4fee\u6539\u533a\u95f4\u5185\uff0c\u5219\u5c06\u8be5\u7ed3\u70b9\u6240\u5b58\u7684\u6570\u52a0\u4e0a(m<\/em>\u00d7\u7ed3\u70b9\u533a\u95f4\u957f\u5ea6)\uff0c\u5e76\u4e14\u5c06\u8be5\u7ed3\u70b9\u7684\u61d2\u6807\u8bb0\u52a0\u4e0am<\/em>\u3002\u5426\u5219\uff0c\u5c06\u8be5\u7ed3\u70b9\u7684\u61d2\u6807\u8bb0\u4e0b\u53d1\uff0c\u5e76\u4e14\u6839\u636e\u533a\u95f4\u7684\u8de8\u533a\u60c5\u51b5\uff0c\u9012\u5f52\u5176\u5de6\u53f3\u5b50\u6811\uff0c\u6700\u540e\u5c06\u7ed3\u70b9\u7684\u5de6\u53f3\u5b50\u7ed3\u70b9\u7684\u6570\u4e0a\u4f20\u56de\u6eaf\u81f3\u8be5\u7236\u7ed3\u70b9\u3002<\/p>\n\n\n\n

void add(int k, int l, int r,int m)<\/p>

{<\/p>

    if (node[k].l >= l && node[k].r <= r)<\/p>

    {<\/p>

       node[k].sum += (node[k].r – node[k].l + 1) * m;<\/p>

       node[k].lazy += m;<\/p>

       return;<\/p>

    }<\/p>

    down(k);<\/p>

    int mid = (node[k].l + node[k].r) \/ 2;<\/p>

    if (l <= mid)<\/p>

       add(k * 2, l, r, m);<\/p>

    if (r > mid)<\/p>

       add(k * 2 + 1, l, r, m);<\/p>

    up(k);<\/p>

}<\/p><\/blockquote>\n\n\n\n

\u516d\u3001\u7ebf\u6bb5\u6811\u7684\u66f4\u591a\u64cd\u4f5c<\/h3>\n\n\n\n

\u9664\u4e86\u533a\u95f4\u67e5\u8be2\u548c\u533a\u95f4\u4fee\u6539\u5916\uff0c\u7ebf\u6bb5\u6811\u8fd8\u53ef\u4ee5\u5b9e\u73b0\u5355\u70b9\u67e5\u8be2\u548c\u5355\u70b9\u4fee\u6539\u3002<\/p>\n\n\n\n

int query_s(int k,int x)<\/p>

{<\/p>

    if (node[k].l == node[k].r)<\/p>

       return node[k].sum;<\/p>

    down(k);<\/p>

    int mid = (node[k].l + node[k].r) \/ 2;<\/p>

    if (x <= mid)<\/p>

       return query_s(k * 2, x);<\/p>

    else<\/p>

       return query_s(k * 2 + 1, x);<\/p>

    return 0;<\/p>

}<\/p>

void add_s(int k,int x,int m)<\/p>

{<\/p>

    if (node[k].l == node[k].r)<\/p>

    {<\/p>

       node[k].sum += m;<\/p>

       return;<\/p>

    }<\/p>

    down(k);<\/p>

    int mid = (node[k].l + node[k].r) \/ 2;<\/p>

    if (x <= mid)<\/p>

       add_s(k * 2, x, m);<\/p>

    else<\/p>

       add_s(k * 2 + 1, x, m);<\/p>

    up(k);<\/p>

}<\/p><\/blockquote>\n\n\n\n

\u4e03\u3001\u5b8c\u6574\u4ee3\u7801<\/h3>\n\n\n\n

#include <stdio.h><\/p>

#define max_n 1000000<\/p>

typedef struct _tNode<\/p>

{<\/p>

    int l;<\/p>

    int r;<\/p>

    int lazy;<\/p>

    int sum;<\/p>

}Node;<\/p>

Node node[4 * max_n + 1];<\/p>

void up(int k)<\/p>

{<\/p>

    node[k].sum = node[k * 2].sum + node[k * 2 + 1].sum;<\/p>

}<\/p>

void down(int k)<\/p>

{<\/p>

    if (node[k].lazy)<\/p>

    {<\/p>

       node[k * 2].lazy += node[k].lazy;<\/p>

       node[k * 2 + 1].lazy += node[k].lazy;<\/p>

       node[k * 2].sum += node[k].lazy * (node[k * 2].r – node[k * 2].l + 1);<\/p>

       node[k * 2 + 1].sum += node[k].lazy * (node[k * 2 + 1].r – node[k * 2 + 1].l + 1);<\/p>

       node[k].lazy = 0;<\/p>

    }<\/p>

}<\/p>

void build(int k,int l,int r)<\/p>

{<\/p>

    if (l > r) <\/p>

       return;<\/p>

    node[k].l = l;<\/p>

    node[k].r = r;<\/p>

    node[k].lazy = 0;<\/p>

    if (l == r)<\/p>

    {<\/p>

       node[k].sum = 1;<\/p>

       return;<\/p>

    }<\/p>

    int mid = (l + r) \/ 2;<\/p>

    build(k * 2, l, mid);<\/p>

    build(k * 2 + 1, mid + 1, r);<\/p>

    up(k);<\/p>

}<\/p>

int query_s(int k,int x)<\/p>

{<\/p>

    if (node[k].l == node[k].r)<\/p>

       return node[k].sum;<\/p>

    down(k);<\/p>

    int mid = (node[k].l + node[k].r) \/ 2;<\/p>

    if (x <= mid)<\/p>

       return query_s(k * 2, x);<\/p>

    else<\/p>

       return query_s(k * 2 + 1, x);<\/p>

    return 0;<\/p>

}<\/p>

void add_s(int k,int x,int m)<\/p>

{<\/p>

    if (node[k].l == node[k].r)<\/p>

    {<\/p>

       node[k].sum += m;<\/p>

       return;<\/p>

    }<\/p>

    down(k);<\/p>

    int mid = (node[k].l + node[k].r) \/ 2;<\/p>

    if (x <= mid)<\/p>

       add_s(k * 2, x, m);<\/p>

    else<\/p>

       add_s(k * 2 + 1, x, m);<\/p>

    up(k);<\/p>

}<\/p>

int query(int k,int l,int r)<\/p>

{<\/p>

    int sum_l = 0, sum_r = 0;<\/p>

    if (node[k].l >= l && node[k].r <= r)<\/p>

       return node[k].sum;<\/p>

    down(k);<\/p>

    int mid = (node[k].l + node[k].r) \/ 2;<\/p>

    if (l <= mid)<\/p>

       sum_l = query(k * 2, l, r);<\/p>

    if (r > mid)<\/p>

       sum_r = query(k * 2 + 1, l, r);<\/p>

    return sum_l + sum_r;<\/p>

}<\/p>

void add(int k, int l, int r,int m)<\/p>

{<\/p>

    if (node[k].l >= l && node[k].r <= r)<\/p>

    {<\/p>

       node[k].sum += (node[k].r – node[k].l + 1) * m;<\/p>

       node[k].lazy += m;<\/p>

       return;<\/p>

    }<\/p>

    down(k);<\/p>

    int mid = (node[k].l + node[k].r) \/ 2;<\/p>

    if (l <= mid)<\/p>

       add(k * 2, l, r, m);<\/p>

    if (r > mid)<\/p>

       add(k * 2 + 1, l, r, m);<\/p>

    up(k);<\/p>

}<\/p>

int main(void)<\/p>

{<\/p>

    int n, m, l, r, s, t;<\/p>

    build(1, 1, max_n);<\/p>

    printf(“n = “);<\/p>

    scanf(“%d”, &n);<\/p>

    printf(“m = “);<\/p>

    scanf(“%d”, &m);<\/p>

    printf(“l = “);<\/p>

    scanf(“%d”, &l);<\/p>

    printf(“r = “);<\/p>

    scanf(“%d”, &r);<\/p>

    if (l > r)<\/p>

    {<\/p>

       l = l ^ r;<\/p>

       r = l ^ r;<\/p>

       l = l ^ r;<\/p>

    }<\/p>

    add(1, l, r, m);<\/p>

    printf(“s = “);<\/p>

    scanf(“%d”, &s);<\/p>

    printf(“t = “);<\/p>

    scanf(“%d”, &t);<\/p>

    if (s > t)<\/p>

    {<\/p>

       s = s ^ t;<\/p>

       t = s ^ t;<\/p>

       s = s ^ t;<\/p>

    }<\/p>

    printf(“sum = %d\\n”, query(1, s, t));<\/p>

    return 0;<\/p>

}<\/p><\/blockquote>\n\n\n\n

\u6d4b\u8bd5\u7ed3\u679c\u5982\u56fe2.7\u6240\u793a\u3002<\/p>\n\n\n\n

\"\"
\u56fe2.7<\/figcaption><\/figure><\/div>\n\n\n\n
\n\n\n\n

3 \u7ebf\u6bb5\u6811\u7684\u5e94\u7528\u2014\u2014POJ 2777 <\/h2>\n\n\n\n

\u94fe\u63a5\uff1ahttp:\/\/poj.org\/problem?id=2777<\/a><\/em><\/p>\n\n\n\n

Count Color<\/strong><\/p>\n\n\n\n
\n Time Limit:<\/strong> 1000MS\n <\/td>\n Memory Limit:<\/strong> 65536K\n <\/td><\/tr>
\n Total Submissions:<\/strong> 54642\n <\/td>\n Accepted:<\/strong> 16407\n <\/td><\/tr><\/tbody><\/table>\n\n\n\n

Description<\/strong><\/p>\n\n\n\n

Chosen Problem Solving and Program design as\nan optional course, you are required to solve all kinds of problems. Here, we\nget a new problem. 
\n
\nThere is a very long board with length L centimeter, L is a positive integer,\nso we can evenly divide the board into L segments, and they are labeled by 1,\n2, … L from left to right, each is 1 centimeter long. Now we have to color\nthe board – one segment with only one color. We can do following two operations\non the board: 
\n
\n1. “C A B C” Color the board from segment A to segment B with color\nC. 
\n2. “P A B” Output the number of different colors painted between\nsegment A and segment B (including). 
\n
\nIn our daily life, we have very few words to describe a color (red, green,\nblue, yellow\u2026), so you may assume that the total number of different colors T\nis very small. To make it simple, we express the names of colors as color 1,\ncolor 2, … color T. At the beginning, the board was painted in color 1. Now\nthe rest of problem is left to your. <\/p>\n\n\n\n

Input<\/strong><\/p>\n\n\n\n

First line of input contains L (1 <= L\n<= 100000), T (1 <= T <= 30) and O (1 <= O <= 100000). Here O\ndenotes the number of operations. Following O lines, each contains “C A B\nC” or “P A B” (here A, B, C are integers, and A may be larger\nthan B) as an operation defined previously.<\/p>\n\n\n\n

Output<\/strong><\/p>\n\n\n\n

Ouput results of the output operation in\norder, each line contains a number.<\/p>\n\n\n\n

Sample Input<\/strong><\/p>\n\n\n\n

2 2 4<\/p>

C 1 1 2<\/p>

P 1 2<\/p>

C 2 2 2<\/p>

P 1 2<\/p><\/blockquote>\n\n\n\n

Sample Output<\/strong><\/p>\n\n\n\n

2<\/p>

1<\/p><\/blockquote>\n\n\n\n

\u9898\u76ee\u5927\u610f<\/strong>\n\u6709\u4e00\u957f\u5ea6\u4e3aL cm\u7684\u677f\uff0c\u5c06\u677f\u5e73\u5747\u5206\u6210L\u5757\uff0c\u6bcf\u4e00\u5757\u7684\u989c\u8272\u9ed8\u8ba4\u4e3a\u989c\u82721\uff0c\u73b0\u5728\u6709T\u79cd\u989c\u6599\uff0c\u4e24\u79cd\u64cd\u4f5c\uff0c\u4e00\u79cd\u662f\u5728\u5757A\u4e0e\u5757B\u4e4b\u95f4\u6d82\u4e0a\u989c\u8272C\uff0c\u53e6\u4e00\u79cd\u662f\u8f93\u51fa\u5757A\u4e0e\u5757B\u4e4b\u95f4\u6709\u591a\u5c11\u79cd\u4e0d\u540c\u7684\u989c\u8272\uff0c\u5171\u6709O\u9879\u64cd\u4f5c\u3002<\/p>\n\n\n\n

\u5206\u6790<\/strong> \u8fd9\u4e00\u9898\u662f\u4e00\u9053\u5178\u578b\u7684\u7ebf\u6bb5\u6811\u7684\u5e94\u7528\u95ee\u9898\u3002\u56e0\u4e3aT\u5c0f\u4e8e30\uff0c\u53ef\u4ee5\u5229\u7528\u4e8c\u8fdb\u5236\u64cd\u4f5c\uff0c\u628a\u533a\u95f4\u7684\u989c\u8272\u5c5e\u6027\u4f5c\u4e3a\u7ebf\u6bb5\u6811\u7ed3\u70b9\u7684\u6807\u8bb0\u6570\u636e\u30021\u8868\u793a\u6709\u8fd9\u79cd\u989c\u8272\uff0c0\u8868\u793a\u6ca1\u6709\u8fd9\u79cd\u989c\u8272\u3002\u4e8c\u8fdb\u5236\u5e8f\u5217\u201c10000001000010000100000000000000\u201d\u8868\u793a\u8be5\u533a\u95f4\u6709\u989c\u82721\u3001\u989c\u82728\u3001\u989c\u827213\u548c\u989c\u827218\u8fd9\u56db\u79cd\u989c\u8272\u3002\u5728\u67e5\u8be2\u67d0\u533a\u95f4\u5757\u542b\u6709\u591a\u5c11\u79cd\u989c\u8272\u65f6\uff0c\u5c06\u533a\u95f4\u7684\u989c\u8272\u5c5e\u6027\u4e8c\u8fdb\u5236\u5e8f\u5217\u4ece\u5de6\u5230\u53f3\u626b\u63cf\uff0c\u5e8f\u5217\u4e2d1\u7684\u4e2a\u6570\u5373\u4e3a\u533a\u95f4\u7684\u989c\u8272\u79cd\u6570\u3002<\/p>\n\n\n\n

\u89e3<\/strong><\/p>\n\n\n\n

#include <stdio.h><\/p>

#define max_n 100000<\/p>

struct _tNode<\/p>

{<\/p>

    int l;<\/p>

    int r;<\/p>

    int c;<\/p>

    int lazy;<\/p>

}node[4 * max_n + 1];<\/p>

void up(int k)<\/p>

{<\/p>

    node[k].c = node[k * 2].c | node[k * 2 + 1].c;<\/p>

}<\/p>

void down(int k)<\/p>

{<\/p>

    if (node[k].lazy)<\/p>

    {<\/p>

       node[k * 2].lazy = node[k * 2 + 1].lazy = node[k].lazy;<\/p>

       node[k * 2].c = node[k * 2 + 1].c = node[k].lazy;<\/p>

       node[k].lazy = 0;<\/p>

    }<\/p>

}<\/p>

void build(int k, int l, int r)<\/p>

{<\/p>

    if (l > r) return;<\/p>

    node[k].l = l;<\/p>

    node[k].r = r;<\/p>

    node[k].lazy = 0;<\/p>

    if (l == r)<\/p>

    {<\/p>

       node[k].c = 1;<\/p>

       return;<\/p>

    }<\/p>

    int m = (l + r) \/ 2;<\/p>

    build(k * 2, l, m);<\/p>

    build(k * 2 + 1, m + 1, r);<\/p>

    up(k);<\/p>

}<\/p>

void paint(int k, int l, int r, int c)<\/p>

{<\/p>

    if (node[k].l >= l && node[k].r <= r)<\/p>

    {<\/p>

       node[k].c = 1 << (c – 1);<\/p>

       node[k].lazy = 1 << (c – 1);<\/p>

       return;<\/p>

    }<\/p>

    down(k);<\/p>

    int m = (node[k].l + node[k].r) \/ 2;<\/p>

    if (l <= m)<\/p>

       paint(k * 2, l, r, c);<\/p>

    if (r > m)<\/p>

       paint(k * 2 + 1, l, r, c);<\/p>

    up(k);<\/p>

}<\/p>

int query(int k, int l, int r)<\/p>

{<\/p>

    int a = 0, b = 0;<\/p>

    if (node[k].l >= l && node[k].r <= r)<\/p>

       return node[k].c;<\/p>

    down(k);<\/p>

    int m = (node[k].l + node[k].r) \/ 2;<\/p>

    if (l <= m)<\/p>

       a = query(k * 2, l, r);<\/p>

    if (r > m)<\/p>

       b = query(k * 2 + 1, l, r);<\/p>

    return a | b;<\/p>

}<\/p>

int main(void)<\/p>

{<\/p>

    int n, t, o;<\/p>

    char op;<\/p>

    int l, r, c;<\/p>

    scanf(“%d %d %d”, &n, &t, &o);<\/p>

    build(1, 1, n);<\/p>

    while (o–)<\/p>

    {<\/p>

       scanf(” %c”, &op);<\/p>

       if (op == ‘C’)<\/p>

       {<\/p>

           scanf(“%d %d %d”, &l, &r, &c);<\/p>

           if (l > r)<\/p>

           {<\/p>

              l = l ^ r;<\/p>

              r = l ^ r;<\/p>

              l = l ^ r;<\/p>

           }<\/p>

           paint(1, l, r, c);<\/p>

        }<\/p>

       else if (op == ‘P’)<\/p>

       {<\/p>

           int total;<\/p>

           int sum = 0;<\/p>

           scanf(“%d %d”, &l, &r);<\/p>

           if (l > r)<\/p>

           {<\/p>

              l = l ^ r;<\/p>

              r = l ^ r;<\/p>

              l = l ^ r;<\/p>

           }<\/p>

           total = query(1, l, r);<\/p>

           while (total)<\/p>

           {<\/p>

              if (total & 1)<\/p>

                  sum++;<\/p>

              total >>= 1;<\/p>

           }<\/p>

           printf(“%d\\n”, sum);<\/p>

       }<\/p>

    }<\/p>

}<\/p><\/blockquote>\n\n\n\n

\u8fd0\u884c\u7ed3\u679c\u5982\u56fe3.1\u6240\u793a\u3002<\/p>\n\n\n\n

\"\"
\u56fe3.1<\/figcaption><\/figure><\/div>\n\n\n\n

\u63d0\u4ea4\u7ed3\u679c\u5982\u56fe3.2\u6240\u793a\u3002<\/p>\n\n\n\n

\"\"
\u56fe3.2<\/figcaption><\/figure><\/div>\n\n\n\n
\n\n\n\n

4 \u603b\u7ed3<\/h2>\n\n\n\n

\u7ebf\u6bb5\u6811\u662f\u4e00\u79cd\u9ad8\u7ea7\u6570\u636e\u7ed3\u6784\uff0c\u5229\u7528\u597d\u7ebf\u6bb5\u6811\uff0c\u53ef\u4ee5\u9ad8\u6548\u7387\u5730\u89e3\u51b3\u533a\u95f4\u95ee\u9898\u548c\u7ebf\u6bb5\u95ee\u9898\u3002\u7ebf\u6bb5\u6811\u7684\u6570\u636e\u7ed3\u6784\u8fd8\u53ef\u4ee5\u4f7f\u7528\u94fe\u5f0f\u7ed3\u6784\uff0c\u51cf\u5c11\u7a7a\u95f4\u5360\u7528\uff0c\u7f3a\u70b9\u662f\u7ed3\u70b9\u7684\u8bbf\u95ee\u6548\u7387\u4f1a\u6709\u6240\u964d\u4f4e\u3002<\/p>\n\n\n\n

\u6587\u4e2d\u7ebf\u6bb5\u6811\u6570\u7ec4\u957f\u5ea6\u4e3a4n<\/em>+1\uff0c\u5728\u5b9e\u9645\u5e94\u7528\u4e0a\u5e76\u7528\u4e0d\u5230\u90a3\u4e48\u591a\uff0c\u8fd9\u9020\u6210\u4e86\u5927\u91cf\u7684\u7a7a\u95f4\u6d6a\u8d39\uff0c\u53ef\u4ee5\u5229\u7528\u6df1\u5ea6\u4f18\u5148\u641c\u7d22\u5e8f\u5217\u8fdb\u884c\u4f18\u5316\uff0c\u4ee3\u66ff\u539f\u6570\u7ec4\u4e0b\u6807\uff0c\u53ef\u4ee5\u628a\u7a7a\u95f4\u5360\u7528\u538b\u7f29\u52302n<\/em>+1\u3002<\/p>\n\n\n\n

\n\n\n\n

5 \u53c2\u8003\u8d44\u6599<\/h2>\n\n\n\n

1.Segment\ntree. https:\/\/en.wikipedia.org\/wiki\/Segment_tree<\/a>. Wikipedia<\/em>.<\/p>\n\n\n\n

2.\u2605\u5fc6\u5148\u2605.\u300a[\u5947\u602a\u4f46\u6709\u7528\u7684\u6570\u636e\u7ed3\u6784]\u7ebf\u6bb5\u6811\u300b. https:\/\/cloud.tencent.com\/developer\/article\/1149069<\/a>. \u817e\u8baf\u4e91<\/em>.<\/p>\n\n\n\n

3. TRTTG<\/a>. \u300a\u6d45\u8c08\u7ebf\u6bb5\u6811\u300b. https:\/\/www.cnblogs.com\/TheRoadToTheGold\/p\/6254255.html<\/a>. \u535a\u5ba2\u56ed<\/em>. <\/p>\n","protected":false},"excerpt":{"rendered":"

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