hdu-2444-二分图判定+最大分配

The Accomodation of Students

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5451    Accepted Submission(s): 2491

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

 题目大意：有n个学生，有m对人是认识的，每一对认识的人能分到一间房，问能否把n个学生分成两部分，每部分内的学生互不认识，而两部分之间的学生认识。如果可以分成两部分，就算出房间最多需要多少间，否则就输出No。

思路：二分判定用染色法，找最大匹配数用匈牙利算法

代码：

#include <iostream>
#include <vector>
#include <cstring>
#include <cstdio>
const int MAX=1e5+5;
using namespace std;
vector<int>mp[MAX];
int d[MAX],n,m,vis[MAX],link[MAX];
int dfs(int x,int f)            //染色法
{
    d[x]=f;
    for(int i=0; i<mp[x].size(); i++)
    {
        if(d[x]==d[mp[x][i]])
            return 0;
        int u=mp[x][i];
        if(d[u]==0)
            if(!dfs(u,-f))
                return 0;
    }
    return 1;
}
int dfs2(int x)             //找最大匹配
{
    for(int i=0; i<mp[x].size(); i++)
    {
        int v=mp[x][i];
        if(!vis[v])
        {
            vis[v]=1;
            if(link[v]==-1||dfs2(link[v]))
            {
                link[v]=x;
                return 1;
            }
        }
    }
    return 0;

}
int find()               //找最大匹配
{ 
    int res=0;
    memset(link,-1,sizeof(link));
    for(int i=1; i<=n; i++)
    {
        memset(vis,0,sizeof(vis));
        if(dfs2(i))
            res++;
    }
    return res;
}
int main()
{
    int a,b;
    while(cin>>n>>m)
    {
        for(int i=1; i<=n; i++)
            if(mp[i].size())
                mp[i].clear();
        for(int i=0; i<m; i++)
        {
            scanf("%d%d",&a,&b);
            mp[a].push_back(b);
            mp[b].push_back(a);
        }
        int flag=1;
        memset(d,0,sizeof(d));
        for(int i=1; i<=n; i++)          //染色
        {
            if(mp[i].size())
                if(!d[i])
                    if(!dfs(i,1))
                    {
                        flag=0;
                        break;
                    }
        }
        if(!flag)
            cout<<"No"<<endl;
        else
        {
            cout<<find()/2<<endl;
        }

    }
}