题意很简单,在一颗树上找最小点覆盖. 将树染成黑白两色,构成一张二分图,然后最大匹配==最小点覆盖即可,所以一次匈牙利就可以求出来了 hdu1054 #include <iostream> #include <stdio.h> #include <string.h> #include <math.h> #include <algorithm> #include <string> #include <queue> #incl…

题目链接:https://vjudge.net/problem/HDU-1054 Strategic Game Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8673    Accepted Submission(s): 4174 Problem Description Bob enjoys playing computer gam…

Strategic Game Bob enjoys playing computer games, especially strategic games, but sometimes he cannot find the solution fast enough and then he is very sad. Now he has the following problem. He must defend a medieval city, the roads of which form a t…

B - Strategic Game Time Limit:10000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1054 Description Bob enjoys playing computer games, especially strategic games, but sometimes he cannot find the solution fast…

Strategic Game Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5034    Accepted Submission(s): 2297 Problem Description Bob enjoys playing computer games, especially strategic games, but somet…

B - 树形dp Crawling in process... Crawling failed Time Limit:2000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u Submit Status Description Bob enjoys playing computer games, especially strategic games, but sometimes he cannot find the sol…

传送门:Strategic Game 题意:用尽量少的顶点来覆盖所有的边. 分析:最小顶点覆盖裸题,最小顶点覆盖=最大匹配数(双向图)/2. #include <cstdio> #include <cstring> #include <string> #include <cmath> #include <iostream> #include <algorithm> #include <queue> #include <…

最小定点覆盖是指这样一种情况: 图G的顶点覆盖是一个顶点集合V,使得G中的每一条边都接触V中的至少一个顶点.我们称集合V覆盖了G的边.最小顶点覆盖是用最少的顶点来覆盖所有的边.顶点覆盖数是最小顶点覆盖的大小. 相应地,图G的边覆盖是一个边集合E,使得G中的每一个顶点都接触E中的至少一条边.如果只说覆盖,则通常是指顶点覆盖,而不是边覆盖. 在二分图中:最大匹配数=最小顶点覆盖数: Bob enjoys playing computer games, especially strategic gam…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1054 思路:树形DP,用二分匹配也能解决 定义dp[root][1],表示以root 为根结点的子树且在root放一个士兵的最优值,那么因为在root已经有士兵了,所以对其孩子来说可以放也可以不放 那么dp[root][1]+=min(dp[son][1],dp[son][0]); dp[root][0]表示以root为根结点的子树且在root没有放士兵的最优值,那么因为在root没有士兵,所以在其…

题意:给你一个图,图里有墙壁和空地,空地可以放置一台机关枪,机关枪可以朝着四个方向发射,子弹不能穿透墙壁,但是射程无限,机关枪会被损坏如果被另一台机关枪的子弹打到,问你最多能放置多少台机关枪: 解题思路:考虑每台机关枪实际能够朝行和列两个方向开火,根据贪心的想法,尽可能不在某行和某列的交点放置,那么如果我们把行和列分成x,y两部分,每行中能够连接的空地当作x的一个顶点,每列中能够连接的空地当作y的一个顶点,问题转换为在二分图中找没有公共顶点的最大边集,也就是二分图最大匹配 代码: #includ…