A new Graph Game Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2360    Accepted Submission(s): 951 Problem Description An undirected graph is a graph in which the nodes are connected by undire…

Mining Station on the Sea Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2997    Accepted Submission(s): 913 Problem Description The ocean is a treasure house of resources and the development o…

奔小康赚大钱 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10760    Accepted Submission(s): 4765 Problem Description 传说在遥远的地方有一个非常富裕的村落,有一天,村长决定进行制度改革:重新分配房子.这可是一件大事,关系到人民的住房问题啊.村里共有n间房间,刚好有n家老百姓,考虑…

Tour Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 2925    Accepted Submission(s): 1407 Problem Description In the kingdom of Henryy, there are N (2 <= N <= 200) cities, with M (M <= 30000…

http://acm.hdu.edu.cn/showproblem.php?pid=4862 选t<=k次,t条路要经过全部的点一次而且只一次. 建图是问题: 我自己最初就把n*m 个点分别放入X集合以及Y集合,再求最优匹配,然后连例子都过不了,并且事实上当时解释不了什么情况下不能得到结果.由于k此这个条件相当于没用上... 建图方法: 1.X集合和Y集合都放入n*m+k个点,X中前n*m个点和Y中前n*m个点之间.假设格子里的值相等.权就是(收益-耗费),不等就是(-耗费),由于要的是最大收益…

Special Fish Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2189    Accepted Submission(s): 826 Problem Description There is a kind of special fish in the East Lake where is closed to campus of…

Going Home Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3299    Accepted Submission(s): 1674 Problem Description On a grid map there are n little men and n houses. In each unit time, every l…

和HDU 3488一样的,只不过要判断一下是否有解. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <vector> using namespace std; + ; const int INF = 0x3f3f3f3f; int n, m; int W[maxn][maxn], lft[maxn]; int slack…

#include<stdio.h> #include<math.h> #include<string.h> #define N 200 #define inf 999999999 int Max(int a,int b ) {     return a>b?a:b; } int Min(int a,int b) {     return a>b?b:a; } int map[N][N],lx[N],ly[N],s[N],t[N],link[N],n,m,ot…

啦啦啦! KM算法是通过给每个顶点一个标号(叫做顶标)来把求最大权匹配的问题转 化为求完备匹配的问题的.设顶点Xi的顶标为A[i],顶点Yi的顶标为B[i],顶点Xi与Yj之间的边权为w[i,j].在算法执行过程中的任一时刻,对于任一条边(i,j), A[i]+B[j]>=w[i,j]始终成立. KM算法的正确性基于以下定理: *  若由二分图中所有满足A[i]+B[j]=w[i,j]的边(i,j)构成的子图(称做相等子图)有完备匹配,那么这个完备匹配就是二分图的最大权匹配. * 这个定理是显然…