Jimmy’s Assignment Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 464    Accepted Submission(s): 224 Problem Description Jimmy is studying Advanced Graph Algorithms at his university. His most…

Description Jimmy is studying Advanced Graph Algorithms at his university. His most recent assignment is to find a maximum matching in a special kind of graph. This graph is undirected, has N vertices and each vertex has degree 3. Furthermore, the gr…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2063 二分匹配最大匹配数简单题,匈牙利算法.学习二分匹配传送门:http://blog.csdn.net/dark_scope/article/details/8880547 #include <iostream> #include <cstdio> #include <cstring> #include <vector> using namespace std…

Description Suppose that we have a square city with straight streets. A map of a city is a square board with n rows and n columns, each representing a street or a piece of wall. A blockhouse is a small castle that has four openings through which to s…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4619 题意: 平面上有一些1×2的骨牌,每张骨牌要么水平放置,要么竖直放置,并且保证同方向放置的骨牌不会相互覆盖.水平放置的牌和竖直放置的牌可能相互覆盖,现在要移去一些牌,使得剩下的牌任何两张都不会相互覆盖,问桌面上最多能剩多少张牌. 分析: 如果把每张牌看作一个结点,则共有两类结点,容易联想到二分图.另外,同方向的牌不会相互覆盖,不同方向的可能相互覆盖,易想到二分图的一个重要性质:同类结点间不会连…

解题报告:有m个女生和n个男生要结成伴坐过山车,每个女生都有几个自己想选择的男生,然后要你确定最多能组成多少对组合. 最裸的一个二分匹配,这是我第一次写二分匹配,给我最大的感受就是看那些人讲的匈牙利算法真不知道他们是怎么写的,看着就觉得很麻烦,好像很难的样子,我这是看第二次,最后还是没看懂,最后实在没办法了,直接看代码了,才发现,原来二分匹配可以用很简单的描述,下面我就用我的语言描述一下: 假如分成两个集合X和Y,我们可以从X 中的第一个元素枚举到最后一个,然后每次枚举的内容就是从X 出发,找这…

Card Game Cheater Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1072    Accepted Submission(s): 564 Problem Description Adam and Eve play a card game using a regular deck of 52 cards. The rule…

Girls and Boys Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6867    Accepted Submission(s): 3083 Problem Description the second year of the university somebody started a study on the romant…

题意:给出每个学生的标号及与其有缘分成为情侣的人的标号,求一个最大集合,集合中任意两个人都没有缘分成为情侣. 分析: 1.若两人有缘分,则可以连一条边,本题是求一个最大集合,集合中任意两点都不相连,即最大独立集问题. 2.最大独立集 = 顶点数 - 最大匹配数(匈牙利算法求解). 3.将一个人拆成两个相同的人进行二分匹配,因此真正的最大匹配数应为得到的最大匹配数/2. #pragma comment(linker, "/STACK:102400000, 102400000") #inc…

//A组n人 B组m人 //最多有多少人匹配 每人仅仅有匹配一次 # include<stdio.h> # include<string.h> # include<algorithm> using namespace std; int n,m,k; int pp[1100][1100],map[1100],vis[1100]; int bfs(int x)//二分匹配模板 { for(int i=1;i<=m;i++)//B组中的人来迎合匹配 { if(!vis[…