hdu2444 二分图的匹配,先判断是否为二分图
来源:互联网 发布:java replace 替换? 编辑:程序博客网 时间:2024/05/17 02:00
http://acm.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 that A and C know each other.
Now you are given all pairs of students who know each other. Your task is to divide the students into two groups so that any two students in the same group don't know each other.If this goal can be achieved, then arrange them into double rooms. Remember, only paris appearing in the previous given set can live in the same room, which means only known students can live in the same room.
Calculate the maximum number of pairs that can be arranged into these double rooms.
Now you are given all pairs of students who know each other. Your task is to divide the students into two groups so that any two students in the same group don't know each other.If this goal can be achieved, then arrange them into double rooms. Remember, only paris appearing in the previous given set can live in the same room, which means only known students can live in the same room.
Calculate the maximum number of pairs that can be arranged into these double rooms.
Input
For each data set:
The first line gives two integers, n and m(1<n<=200), indicating there are n students and m pairs of students who know each other. The next m lines give such pairs.
Proceed to the end of file.
The first line gives two integers, n and m(1<n<=200), indicating there are n students and m pairs of students who know each other. The next m lines give such pairs.
Proceed to the end of file.
Output
If these students cannot be divided into two groups, print "No". Otherwise, print the maximum number of pairs that can be arranged in those rooms.
Sample Input
4 41 21 31 42 36 51 21 31 42 53 6
Sample Output
No3
Source
2008 Asia Harbin Regional Contest Online
/**hdu2444 二分图的匹配,先判断是否为二分图题目大意:给定一个图,判断是否为二分图如果是请求出最大匹配数解题思路:对于是不是二分图:所有的点可以分到两部分,每一部分的点之间没有联系,我们可以用dfs搜索就知道了。 至于二分图的匹配要用到匈牙利算法*/#include <string.h>#include <stdio.h>#include <algorithm>#include <vector>#include <iostream>using namespace std;const int maxn=202;vector <int>vec[maxn];int linker[maxn];bool used[maxn];int n,m;int matchs[maxn],cnt[maxn];bool dfs(int u){ for(int i=0;i<vec[u].size();i++) { int v=vec[u][i]; if(!used[v]) { used[v]=true; if(linker[v]==-1||dfs(linker[v])) { linker[v]=u; return true; } } } return false;}int hungary(){ int res=0; int u; memset(linker,-1,sizeof(linker)); for(int u=1;u<=n;u++) { memset(used,false,sizeof(used)); if(dfs(u)) res++; } return res;}bool judge(int x,int y){ int i; for(int i=0;i<vec[x].size();i++) { if(cnt[vec[x][i]]==0) { cnt[vec[x][i]]=y^1; matchs[vec[x][i]]=true; if(!judge(vec[x][i],y^1))return false; } else if(cnt[vec[x][i]]==y) return false; } return true;}bool matched()///判断是否为二分图{ memset(matchs,false,sizeof(matchs)); for(int i=1;i<=n;i++) { if(vec[i].size()&&!matchs[i]) { memset(cnt,0,sizeof(cnt)); cnt[i]=1; matchs[i]=true; if(!judge(i,1)) return false; } } return true;}int main(){ while(~scanf("%d%d",&n,&m)) { for(int i=1;i<=n;i++) { if(vec[i].size()) vec[i].clear(); } while(m--) { int u,v; scanf("%d%d",&u,&v); vec[u].push_back(v); vec[v].push_back(u); } if(matched()) printf("%d\n",hungary()/2); else printf("No\n"); } return 0;}
0 0
- hdu2444 二分图的匹配,先判断是否为二分图
- hdu2444二分图判断+最大匹配
- hdu2444 判断二分图+最大匹配
- 【hdu2444】判断二分图+求最大匹配
- HDU2444-判断是否能构成二分图
- hdu2444(二分图判断+二分图最大匹配)
- hdu2444 二分图匹配(判断二分图)
- hdu2444二分图最大匹配
- hdu2444 二分图判定+匹配
- hdu2444(判断二分图)
- hdu2444 The Accomodation of Students (二分图判断+最大匹配)
- hdu2444 The Accomodation of Students【二分图判断+最大匹配】
- hdu2444二分图判定 + 最大匹配
- HDU2444 【二分图判定+最大匹配】
- HDU 2444 The Accomodation of Students 判断是否为二分图,二分图的最大匹配
- hdu2444(二分图判定+二分图匹配)
- Hdu2444二分图
- HDU2444(二分图)
- android子线程中更新UI的方法
- 1.搭起环境
- 日经社説 20150304 献金疑惑を招いた制度の不備
- html+js 左移右移
- 信息提醒:AlertView和UIAlertController
- hdu2444 二分图的匹配,先判断是否为二分图
- 日经社説 20150304 多様なサービスを競い合える銀行法制に
- Difference Between Paging and Swapping
- ubuntu下如何用命令行运行deb安装包
- git: command not found
- android错误之android.content.res.Resources$NotFoundException:
- C++学习:string用法
- bootstrap 栅格系统
- 社説 20150304 発送電分離案 安定供給体制の確保が前提だ