hdu2444 二分图的匹配，先判断是否为二分图

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.

 

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.

 

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 4
1 2
1 3
1 4
2 3
6 5
1 2
1 3
1 4
2 5
3 6

 

Sample Output

No
3

 

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;
}