Marriage Match IV Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6230    Accepted Submission(s): 1804 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3416 Description: Do not sincere non-interfe…

http://acm.hdu.edu.cn/showproblem.php?pid=3416 题意:给出n个点m条边,边信息分别是两个端点和一个费用,再给出一个起点和一个终点,问从起点到终点的完全不相同的最短路径有多少条.(即走过的边不能在走过了). 思路:因为是在网络流专题里面,所以一开始以为先用SPFA跑一个最小费用出来,然后再用最小费用最大流(然而是最小费用最大流是满足最大流的前提下再考虑最小费用的,很明显是行不通的).后来想要保证路径不重复,就跑完一次最短路就删除路径(好像也是行不通).…

Marriage Match IV 题目链接: http://acm.hust.edu.cn/vjudge/contest/122685#problem/Q Description Do not sincere non-interference. Like that show, now starvae also take part in a show, but it take place between city A and B. Starvae is in city A and girls a…

HDU 3416 Marriage Match IV (最短路径,网络流,最大流) Description Do not sincere non-interference. Like that show, now starvae also take part in a show, but it take place between city A and B. Starvae is in city A and girls are in city B. Every time starvae can…

题目链接:https://vjudge.net/problem/HDU-3416 Marriage Match IV Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4710    Accepted Submission(s): 1412 Problem Description Do not sincere non-interferenc…

hdu 3416 Marriage Match IV Description Do not sincere non-interference. Like that show, now starvae also take part in a show, but it take place between city A and B. Starvae is in city A and girls are in city B. Every time starvae can get to city B a…

Q - Marriage Match IV Do not sincere non-interference. Like that show, now starvae also take part in a show, but it take place between city A and B. Starvae is in city A and girls are in city B. Every time starvae can get to city B and make a data wi…

Marriage Match IV http://acm.hdu.edu.cn/showproblem.php?pid=3416 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6081    Accepted Submission(s): 1766 Problem Description Do not sincere non-inter…

http://acm.hdu.edu.cn/showproblem.php?pid=3081 题意:有n个男生n个女生,他们只有没有争吵或者女生a与男生A没有争吵,且女生b与女生a是朋友,因此女生b也可以和男生A过家家(具有传递性).给出m个关系,代表女生a和男生b没有争吵过.给出k个关系,代表女生a与女生b是好朋友.每一轮过家家之后,女生只能选择可以选择并且没选过的男生过家家,问游戏能进行几轮. 思路:因为n<=100,因此支持O(n^3)的算法,挺容易想到是一个二分图匹配的.(出现在我的网络…

题意:给你n个点,m条边的图(有向图,记住一定是有向图),给定起点和终点,问你从起点到终点有几条不同的最短路 分析:不同的最短路,即一条边也不能相同,然后刚开始我的想法是找到一条删一条,然后光荣TLE 搜了一下,然后看到网络流,秒懂,就是把所有在最短路上的边重新建一张图,起点到终点的最大流就是解 怎么找到最短路径上的边呢? 在进行dij的时候,每次松弛操作,会更新一个点到起点的最短距离,然后记录一下,对于每一个点 记录有多少点可以走到他可以得到的最短距离,就是记录所有可能的前驱(这里前驱的话,记…