The Accomodation of Students Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description There are a group of students. Some of them may know each other, while others don't. For example, A and B know each o…

The Accomodation of Students Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3775    Accepted Submission(s): 1771 Problem Description There are a group of students. Some of them may know each ot…

The Accomodation of StudentsTime Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8939    Accepted Submission(s): 3925 Problem DescriptionThere are a group of students. Some of them may know each othe…

http://acm.hdu.edu.cn/showproblem.php?pid=2444 The Accomodation of Students Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 2444 Description There are a group of students. Some of them may know…

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=2444 Problem Description There are a group of students. Some of them may know each other, while others don't. For example, A and B know each other, B and C know each other. But this may not imply t…

此题就是求最大匹配.不过需要判断是否构成二分图.判断的方法是人选一点标记为红色(0),与它相邻的点标记为黑色(1),产生矛盾就无法构成二分图.声明一个vis[],初始化为-1.通过深搜,相邻的点不满足异或关系就结束.如果没被标记,就标记为相邻点的异或. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<queue> using na…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=2444 题意:首先判断所有的人可不可以分成互不认识的两部分.如果可以分成 ,则求两部分最多相互认识的对数. 思路:二分图最大匹配问题.先BFS判断是否为二分图,再用匈牙利算法算最大匹配量 关于匈牙利算法:https://blog.csdn.net/CillyB/article/details/55511666 代码: #include<iostream> #include<cstdio>…

这是一个基础的二分图,题意比较好理解,给出n个人,其中有m对互不了解的人,先让我们判断能不能把这n对分成两部分,这就用到的二分图的判断方法了,二分图是没有由奇数条边构成环的图,这里用bfs染色法就可以判断,其次让我们求分在两部分的最大对数,这就是二分图的最大匹配问题,这里数据只有200,所以匈牙利算法求蹭广路径的办法可以解决这个问题,也相对比较容易编写. 另外一开始我链式前向星的数组开小了,G++居然返回超时,后来换了C++才RE,想到数组越界的问题,不得不说这些编译器真傲娇啊- #includ…

[题目链接]:pid=2444">click here~~ [题目大意]: 给出N个人和M对关系,表示a和b认识,把N个人分成两组,同组间随意俩人互不认识.若不能分成两组输出No,否则输出两组间俩人互相认识的对数 [解题思路]:   先推断是否能构成二分图,推断二分图用交叉染色法:从某个未染色的点出发把此点染成白色,该点周围的点染成黑色.黑色周围的又染成白色.若走到某个点已经染色,而且它相邻点的颜色与它一样则不是二分图,能够这样理解,染白色既增加X集合,黑色既增加Y集合,若某个点即是X集合…

There are a group of students. Some of them may know each other, while others don't. For example, A and B know each other, B and C know each other. But this may not imply that A and C know each other. Now you are given all pairs of students who know…