/* HDU 6073 - Matching In Multiplication [ 图论 ] | 2017 Multi-University Training Contest 4 题意: 定义一张二分图,U中每个节点和V中两个节点连边 完美匹配的权值为该匹配所有边的权值相乘 求所有完美匹配的权值之和 分析: 可以发现有些V中的点只能连唯一的U中的点 按拓扑排序思路将这些全部处理掉后,剩下的点构成一个个环 每个环有两种连线方式,间隔取边权 */ #include <bits/stdc++.h>…

Matching In Multiplication Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 1389    Accepted Submission(s): 423 Problem Description In the mathematical discipline of graph theory, a bipartite g…

Matching In Multiplication Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 1127    Accepted Submission(s): 325 Problem Description In the mathematical discipline of graph theory, a bipartite g…

Matching In Multiplication Problem DescriptionIn the mathematical discipline of graph theory, a bipartite graph is a graph whose vertices can be divided into two disjoint sets U and V (that is, U and V are each independent sets) such that every edge…

题目链接 Problem Description In the mathematical discipline of graph theory, a bipartite graph is a graph whose vertices can be divided into two disjoint sets U and V (that is, U and V are each independent sets) such that every edge connects a vertex in…

http://acm.hdu.edu.cn/showproblem.php?pid=6073 题意:有个二分图,左边和右边的顶点数相同,左边的顶点每个顶点度数为2.现在有个屌丝理解错了最佳完美匹配,它以为最佳完美匹配是边权的乘积了,现在要你计算所有这种最佳完美匹配的边权乘积和.保证至少存在一个完美匹配. 思路: 这道题目虽然打着二分图的幌子,但其实吧,根本就不是二分图,哎. 对于右边的点来说,如果它的度数为1,那么与它匹配的点肯定是确定的,所以我们先通过拓扑排序来计算出所有确定的匹配.去除这些点…

/* HDU 6162 - Ch’s gift [ LCA,线段树 ] | 2017 ZJUT Multi-University Training 9 题意: N节点的树,Q组询问 每次询问s,t两节点之间的路径上点权值在[a,b]之间的点权总和 分析: 求出每个询问的LCA,然后离线 按dfs顺序更新树状数组,即某点处树状数组中存的值为其所有祖先节点的值 每个点处对答案的贡献为: 当其为第 i 个 lca 时, ans[i] -= 2 * query(a,b) , 再特判该节点 当其为第 i…

/* HDU 6170 - Two strings [ DP ] | 2017 ZJUT Multi-University Training 9 题意: 定义*可以匹配任意长度,.可以匹配任意字符,问两串是否匹配 分析: dp[i][j] 代表B[i] 到 A[j]全部匹配 然后根据三种匹配类型分类讨论,可以从i推到i+1 复杂度O(n^2) */ #include <bits/stdc++.h> using namespace std; const int N = 2505; int t;…

Matching In Multiplication 题解: 首先如果一个点的度数为1,那么它的匹配方案是固定的,继而我们可以去掉这一对点.通过拓扑我们可以不断去掉所有度数为1的点. 那么剩下的图中左右各有m个点,每个点度数都不小于2,且左边每个点度数都是2,而右侧总度数是2m,因此右侧只能是每个点度数都是2.这说明这个图每个连通块是个环,在环上间隔着取即可,一共两种方案. 时间复杂度O(n). 比赛的时候已经想出做法了,然而实现的太慢了,时间不够了,最后看了题解才想到,哦,原来隔着取就好了,我…

Matching In Multiplication Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 787    Accepted Submission(s): 222 Problem Description In the mathematical discipline of graph theory, a bipartite gr…