HDU2444 ：The Accomodation of Students(二分图染色+二分图匹配)

The Accomodation of Students

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9477    Accepted Submission(s): 4165

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2444

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

题意：

n个学生,m对互相认识彼此，问能不能把互相认识的分到同一个房间，不能的话输出No，能的话输出最大对数。

 

题解：

考虑能的情况，显然是二分图的最大匹配，A和B一个房间，可以看作A,B以选中，A,B自然不能和其它人在一个房间，其余的也同理。

题干中说的是分成两个group，每一组中的人互不认识，这样就可以构成一个二分图。如果存在二分图，那么就必然有最大匹配。

如果同一组中有认识的人，那么就不能构成二分图，所以还需要二分图的判断（二分图染色）。

注意一点细节：由于要二分图染色，所以连的是双向边，最后求最大匹配的时候求了两次。

 

代码如下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int  N = 205;
int link[N][N],match[N],check[N],color[N];
int n,m,ans=0;

inline void init(){
    ans=0;memset(color,-1,sizeof(color));
    memset(link,0,sizeof(link));memset(match,-1,sizeof(match));
}

inline int dfs(int x){
    for(int i=1;i<=n;i++){
        if(link[x][i] && !check[i]){
            check[i]=1;
            if(match[i]==-1 || dfs(match[i])){
                match[i]=x;
                return 1;
            }
        }
    }
    return 0;
}
inline bool ok(int x){ //二分图染色 
    for(int i=1;i<=n;i++){
        if(link[x][i]){
            if(color[i]==-1){
                color[i]=1-color[x];
                if(!ok(i)) return false;
            }else if(color[i]==color[x]) return false ;
        }
    }
    return true;
}
/*dfs实现 ，连通图直接调用ok(1,0) 
inline bool ok(int x,int c){ 
    color[x]=c;
    for(int i=1;i<=n;i++){
        if(link[x][i]){
            if(color[i]==-1){
                if(!ok(i,1-color[x])) return false;
            }else if(color[i]==color[x]) return false ;
        }
    }
    return true;
}
*/ 
int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        init();
        for(int i=1,a,b;i<=m;i++){
            scanf("%d%d",&a,&b);
            link[a][b]=1;
            link[b][a]=1;
        }
        bool flag=false ;
        for(int i=1;i<=n;i++){
            if(color[i]==-1){
                color[i]=0;
                if(!ok(i)){
                    flag=true;break;
                }
            }
        }
        if(flag){
            puts("No");continue;
        }
        for(int i=1;i<=n;i++){
            memset(check,0,sizeof(check));
            if(dfs(i)) ans++;
        }
        printf("%d\n",ans/2);
    }
    return 0;
}