AC日记——Destroying The Graph poj 2125】的更多相关文章

Destroying The Graph Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8356   Accepted: 2696   Special Judge Description Alice and Bob play the following game. First, Alice draws some directed graph with N vertices and M arcs. After that B…
青蛙的约会 POJ - 1061   思路: 扩展欧几里得: 设青蛙们要跳k步,我们可以得出式子 m*k+a≡n*k+b(mod l) 式子变形得到 m*k+a-n*k-b=t*l (m-n)*k-t*l=b-a 然后,exgcd函数求出k 然后输出刚刚大于0的k即可   来,上代码: #include <cmath> #include <cstdio> #include <cstring> #include <iostream> using namespa…
Destroying The Graph http://poj.org/problem?id=2125 Time Limit: 2000MS   Memory Limit: 65536K           Description Alice and Bob play the following game. First, Alice draws some directed graph with N vertices and M arcs. After that Bob tries to dest…
Destroying The Graph Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8198   Accepted: 2635   Special Judge Description Alice and Bob play the following game. First, Alice draws some directed graph with N vertices and M arcs. After that B…
网络流/二分图最小点权覆盖 果然还是应该先看下胡伯涛的论文…… orz proverbs 题意: N个点M条边的有向图,给出如下两种操作.删除点i的所有出边,代价是Ai.删除点j的所有入边,代价是Bj.求最后删除图中所有的边的最小代价. 其实就是二分图最小点权覆盖. 定义:从x或者y集合中选取一些点,使这些点覆盖所有的边,并且选出来的点的权值尽可能小. 题解: 拆点.n个点拆成2n个点(左右各n个,i与(i+n)对应,之间连容量INF的边),S和i连容量为Ai的边,(i+n)与T之间连容量为Bi…
Destroying The Graph   Description Alice and Bob play the following game. First, Alice draws some directed graph with N vertices and M arcs. After that Bob tries to destroy it. In a move he may take any vertex of the graph and remove either all arcs…
Destroying The Graph Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8311   Accepted: 2677   Special Judge Description Alice and Bob play the following game. First, Alice draws some directed graph with N vertices and M arcs. After that B…
                                                      Destroying The Graph Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8158   Accepted: 2620   Special Judge Description Alice and Bob play the following game. First, Alice draws some d…
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Destroying The Graph 构图思路: 1.将所有顶点v拆成两个点, v1,v2 2.源点S与v1连边,容量为 W- 3.v2与汇点连边,容量为 W+ 4.对图中原边( a, b ), 连边 (a1,b2),容量为正无穷大 则该图的最小割(最大流)即为最小花费. 简单证明: 根据ST割集的定义,将顶点分成两个点集.所以对于原图中的边(a,b),转换成 S->a1->b2->T. 则此时路径必定存在 一条割边,因为a1->b2为无穷大,所以割边必定是 S->a1…