题意是有n座城市,n号城市不想让1号城市可达n号,每条道路有一条毁坏的代价,1号还可以修一条不能毁坏的道路,求n号城市所需的最小代价最大是多少. 毁坏的最小代价就直接求一遍最大流,就是最小割了.而可以修一条不能毁坏的路,需要枚举的这些边就是源集中的点到汇集中的点,我之前的做法是直接找出所有的割边,再枚举每一条割边,在这两个点上加一个流量为无穷大的边再去求最大流,实际上这样连样例的第二个数据都过不了,因为图不保证是连通的,那就有一些源集中的点和汇集中的点本来就不联通,就不存在这样的割边了.发现这个…

There is a war Problem Description       There is a sea.       There are N islands in the sea.       There are some directional bridges connecting these islands.       There is a country called Country One located in Island 1.       There is another…

There is a war Time Limit: 1000ms Memory Limit: 32768KB This problem will be judged on HDU. Original ID: 2435 64-bit integer IO format: %I64d      Java class name: Main There is a sea.There are N islands in the sea.There are some directional bridges…

Friendship Time Limit: 2000MS   Memory Limit: 20000K Total Submissions: 8025   Accepted: 2224 Description In modern society, each person has his own friends. Since all the people are very busy, they communicate with each other only by phone. You can…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1676 对顶点i,j,起点s=1,终点t=n,可以认为题意要求一组01矩阵use[i][j],使得aveCost=sigma(use[i][j]*cost[i][j])/sigma(use[i][j])最小,且{(i,j)|use[i][j]==1}是图的S-T割 定义F(e)=min(sigma(use[i][j]*(cost[i][j]-a))),明显,F(e)是目标式的变…

Problem Description Consider a network G=(V,E) with source s and sink t. An s-t cut is a partition of nodes set V into two parts such that s and t belong to different parts. The cut set is the subset of E with all edges connecting nodes in different…

要选出一些点,这些点之间没有相邻边且要求权值之和最大,求这个权值 分析:二分图带权最大独立集. 用最大流最小割定理求解.其建图思路是:将所有格点编号,奇数视作X部,偶数视作Y部,建立源点S和汇点T, S向X部的点建边,Y部向T建边,容量为该点权值. 相邻的一对点(肯定是一奇一偶),由X中的点向Y中的点建边,容量为正无穷. 最后跑出最大流,|带权最大独立集| = |点权之和| - |最小割| = |点权之和| - |最大流| #include<iostream> #include<cstr…

Problem Description Consider a network G=(V,E) with source s and sink t . An s-t cut is a partition of nodes set V into two parts such that s and t belong to different parts. The cut set is the subset of E with all edges connecting nodes in different…

Hold Your Hand Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/102400 K (Java/Others)Total Submission(s): 169    Accepted Submission(s): 38 Problem Description She walks in beauty, like the night of cloudless climes and starry skies. An…

题意:有一个区域,有'.'的陆地,'D'的深海域,'E'的浅海域.其中浅海域可以填充为陆地.这里的陆地区域不联通,并且整个地图都处在海洋之中.问填充一定浅海域之后所有岛屿的最长的海岸线之和. 解法:最小割.从"分隔"陆地和海域可以想到"割"的概念,然后我们先不考虑浅海域,要深海域和陆地的对数尽量大,也就是相同的地理构造对数尽量小.      于是我们建立一个二分图,深海域和陆地都分列两侧,对于相邻的一个在左,一个在右,也就是"行+列"是奇数的放左…