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摘要: 背包第K大。。。。。。。Bone Collector IITime Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1448 Accepted Submission(s): 729Problem DescriptionThe title of ... 阅读全文
posted @ 2013-07-23 00:24 码代码的猿猿 阅读(184) 评论(0) 推荐(0)
摘要: #include #include #include using namespace std;const int CHAR=26,MAXN=100000;struct Trie{ int tot,root,child[MAXN][CHAR]; bool flag[MAXN]; Trie() { memset(child[1],0,sizeof(child[1]... 阅读全文
posted @ 2013-07-22 05:40 码代码的猿猿 阅读(190) 评论(0) 推荐(0)
摘要: 首先,在字符串s中,用rad表示第i个字符的回文半径,即rad尽可能大,且满足:s[i-rad,i-1]=s[i+1,i+rad]很明显,求出了所有的rad,就求出了所有的长度为奇数的回文子串.至于偶数的怎么求,最后再讲.假设现在求出了rad[1..i-1],现在要求后面的rad值,并且通过前面的操作,得知了当前字符i的rad值至少为j.现在通过试图扩大j来扫描,求出了rad.再假设现在有个指针k... 阅读全文
posted @ 2013-07-22 01:56 码代码的猿猿 阅读(208) 评论(0) 推荐(0)
摘要: manacher。。。。。。。。。。。。。。。。。。。。最长回文Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5240 Accepted Submission(s): 1784Problem Description给出一个只由小写英... 阅读全文
posted @ 2013-07-22 01:23 码代码的猿猿 阅读(153) 评论(0) 推荐(0)
摘要: http://www.matrix67.com/blog/archives/115#include #include #include #include #include using namespace std;vector kmp(string patten,string text){ int n=patten.size(); vector next(n+1,0); for(i... 阅读全文
posted @ 2013-07-21 22:15 码代码的猿猿 阅读(168) 评论(0) 推荐(0)
摘要: Pascal's TravelsTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1296 Accepted Submission(s): 570Problem DescriptionAn n x n game board is pop... 阅读全文
posted @ 2013-07-20 07:45 码代码的猿猿 阅读(181) 评论(0) 推荐(0)
摘要: prim。。。。。畅通工程Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 12261 Accepted Submission(s): 5005Problem Description省政府“畅通工程”的目标是使全省任何两个村庄间都可以实... 阅读全文
posted @ 2013-07-19 20:01 码代码的猿猿 阅读(152) 评论(0) 推荐(0)
摘要: 还是kruskal。。。。还是畅通工程Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 19402 Accepted Submission(s): 8618Problem Description某省调查乡村交通状况,得到的统计表中列出了... 阅读全文
posted @ 2013-07-19 18:56 码代码的猿猿 阅读(161) 评论(0) 推荐(0)
摘要: 其实就是并查集的应用,先将邻接矩阵换成边集,再按权值由小到大排序,最后用Unionset构造最小生成树。。。。带有优化的并查集=。=int find_father(int x){ int tot=0; while(x!=father[x]) { save[tot++]=x; x=father[x]; } for(int i=0;i<tot;... 阅读全文
posted @ 2013-07-18 08:45 码代码的猿猿 阅读(248) 评论(0) 推荐(0)
摘要: MST的kruska算法。。。。Constructing RoadsTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10595 Accepted Submission(s): 3940Problem DescriptionThere ... 阅读全文
posted @ 2013-07-18 08:42 码代码的猿猿 阅读(193) 评论(0) 推荐(0)
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