Muddy Fields Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7340   Accepted: 2715 Description Rain has pummeled the cows' field, a rectangular grid of R rows and C columns (1 <= R <= 50, 1 <= C <= 50). While good for the grass, t…

POJ 2226 Muddy Fields 题目链接 题意:给定一个图,要求用纸片去覆盖'*'的位置.纸片能够重叠.可是不能放到'.'的位置,为最少须要几个纸片 思路:二分图匹配求最小点覆盖.和放车那题基本一样.就是注意要预处理一下行列,把连续横的'*'当成一行,竖的'*'当成一列,建图跑最小点覆盖就可以 代码: #include <cstdio> #include <cstring> #include <vector> #include <algorithm&g…

http://poj.org/problem?id=2226 Muddy Fields Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7078   Accepted: 2622 Description Rain has pummeled the cows' field, a rectangular grid of R rows and C columns (1 <= R <= 50, 1 <= C <= 5…

http://poj.org/problem?id=2226 Muddy Fields Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9022   Accepted: 3348 Description Rain has pummeled the cows' field, a rectangular grid of R rows and C columns (1 <= R <= 50, 1 <= C <= 5…

  Description Rain has pummeled the cows' field, a rectangular grid of R rows and C columns (1 <= R <= 50, 1 <= C <= 50). While good for the grass, the rain makes some patches of bare earth quite muddy. The cows, being meticulous grazers, don'…

Description Rain has pummeled the cows' field, a rectangular grid of R rows and C columns (1 <= R <= 50, 1 <= C <= 50). While good for the grass, the rain makes some patches of bare earth quite muddy. The cows, being meticulous grazers, don't…

/* 题意:用木板盖住泥泞的地方,不能盖住草.木板任意长!可以重叠覆盖! '*'表示泥泞的地方,'.'表示草! 思路: 首先让我们回忆一下HDU 2119 Matrix这一道题,一个矩阵中只有0, 1,然后让我们通过选择一行,或者 是一列将其所在行的或者所在列的 1全部删掉,求出最少需要几步? 这道题的思路就是:将行标 和 列标值为1的建立一条边!通过匈牙利算法可以得到这个二分图的最大匹配数 最大匹配数==最小顶点覆盖数!最小顶点覆盖就是用最少的点覆盖了这个二分图的所有的边,然后去掉这些 最小覆…

[题目链接] http://poj.org/problem?id=2226 [题目大意] 给出一张图,上面有泥和草地,有泥的地方需要用1*k的木板覆盖, 有草地的地方不希望被覆盖,问在此条件下需要的最少木板数目 [题解] 我们将四周都当成草地,将每块草地拆成横向点和纵向点 对于每一块泥地,我们将其向左和向上遇到的第一块草地连线, 对于这个图的最小点覆盖就是答案,而最小点覆盖等于二分图的最大匹配, 因此我们做一遍最大匹配即可 [代码] #include <cstdio> #include <…

题目地址:http://poj.org/problem?id=2226 二分图的题目关键在于建图.因为“*”的地方只有两种木板覆盖方式:水平或竖直,所以运用这种方式进行二分.首先按行排列,算出每个"*"的序号xi,再按列排序,算出序号yi. 从X集合向Y集合连边.G[xi][yi]=1; 然后就是求二分图的最小顶点覆盖.因为二分图最小点覆盖=最大匹配数.所以匈牙利算法求一下最大匹配就可以了. #include<cstdio> #include<iostream>…

Rain has pummeled the cows' field, a rectangular grid of R rows and C columns (1 <= R <= 50, 1 <= C <= 50). While good for the grass, the rain makes some patches of bare earth quite muddy. The cows, being meticulous grazers, don't want to get…