hdu 2444(二分图最大匹配)

先判断是否为二分图 ， dfs 对每个点染色， 如果不会出现矛盾就说明是二分图。

然后就是匈牙利直接水过了。。。

 

The Accomodation of Students

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1345    Accepted Submission(s): 661

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

 

 

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gaojie

 

 

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;

int n,m;
int g[202][202];
int flag;
int mark[202],pre[202];
int save[202],save1[202];
int cnt,cnt1;

void pdfs(int s,int f)
{
    if(flag==1) return ;
    if(f==0)
        save[cnt++]=s;
    else save1[cnt1++]=s;
    mark[s]=f;
    for(int i=1;i<=n;i++)
    {
        if(g[s][i]==0) continue;
        if(mark[i]==-1)
            pdfs(i,f^1);
        else 
        {
            if(mark[i] != f^1 )
            {
                flag=1;
                return ;
            }
        }
    }
}
int dfs(int s)
{
    for(int i=0;i<cnt1;i++)
    {
        if(mark[i]==1 || g[ save[s] ][ save1[i] ]==0) continue;
        mark[i]=1;
        if( pre[i]==-1 || dfs(pre[i])==1 )
        {
            pre[i]=s;
            return 1;
        }
    }
    return 0;
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        cnt=0; cnt1=0;
        flag=0;
        memset(g,0,sizeof(g));
        for(int i=0;i<m;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            g[x][y]=g[y][x]=1;
        }
        memset(mark,-1,sizeof(mark));
        for(int i=1;i<=n;i++)
        {
            if(mark[i]==-1)
            {
                //save[cnt++]=i;
                pdfs(i,0);
                if(flag==1) break;
            }
        }
        if(flag==1) 
        {
            printf("No\n");
            continue;
        }
        memset(pre,-1,sizeof(pre));
        int sum=0;
        for(int i=0;i<cnt;i++)
        {
            memset(mark,0,sizeof(mark));
            sum+=dfs(i);
        }
        printf("%d\n",sum);
    }
    return 0;
}