hdu 2444 The Accomodation of Students 判断二分图+二分匹配

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 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

用染色法，即从其中一个顶点开始，将跟它邻接的点染成与其不同的颜色，如果邻接的点有相同颜色的，则说明不是二分图，每次用bfs遍历即可。

二分匹配：套版子；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x) cout<<"bug"<<x<<" "<<endl;
const int N=2e2+10,M=1e6+10,inf=2e9+10,mod=1e9+7;
const ll INF=1e18+10;
int n,m;
int mp[N][N];
int linker[N];
bool used[N];
bool dfs(int a)
{
    for(int i=0;i<n;i++)
      if(mp[a][i]&&!used[i])
      {
          used[i]=true;
          if(linker[i]==-1||dfs(linker[i]))
          {
              linker[i]=a;
              return true;
          }
      }
      return false;
}
int hungary()
{
    int result=0;
    memset(linker,-1,sizeof(linker));
    for(int i=0;i<n;i++)
    {
        memset(used,0,sizeof(used));
        if(dfs(i))  result++;
    }
    return result;
}
queue<int>q;
int color[N];
int bfs(int s)
{
    while(!q.empty())q.pop();
    q.push(s);
    while(!q.empty())
    {
        int x=q.front();
        q.pop();
        for(int i=0;i<n;i++)
        {
            if(mp[x][i])
            {
                if(color[i]==-1)
                    color[i]=color[x]^1,q.push(i);
                else if(color[i]==color[x])
                    return 0;
            }
        }
    }
    return 1;
}
int check()
{
    memset(color,-1,sizeof(color));
    for(int i=0;i<n;i++)
    {
        if(color[i]==-1)
        {
            color[i]=1;
            if(!bfs(i))return 0;
        }
    }
    return 1;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(mp,0,sizeof(mp));
        for(int i=1;i<=m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            mp[u-1][v-1]=1;
        }
        if(!check())printf("No\n");
        else
        {
            int cnt=hungary();
            printf("%d\n",cnt);
        }
    }
    return 0;
}