hdu 2444 The Accomodation of Students

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7091    Accepted Submission(s): 3169

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

 

 

题意：给定m对认识关系，不可传递

问能否将n的人分成两组，人员互相不认识。

不能分成两组输出No

否则输出最大匹配数

二分图染色+匹配

屠龙宝刀点击就送

#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#define N 300
using namespace std;
vector<int>G[N];
bool vis[N],flag;
int Map[N][N],match[N],col[N],cnt,n,m;
int dfs(int x)
{
    for(int i=1;i<=n;++i)
    {
        if(Map[x][i]&&!vis[i])
        {
            vis[i]=1;
            if(!match[i]||dfs(match[i]))
            {
                match[i]=x;
                return 1;
            }
        }
    }
    return 0;
}
bool rs(int x)
{
    queue<int>q;
    q.push(x);
    col[x]=0;
    for(int now;!q.empty();)
    {
        now=q.front();
        q.pop();
        for(int i=0;i<G[now].size();++i)
        {
            int v=G[now][i];
            if(col[v]!=-1) {if(col[now]==col[v]) {flag=1;return true;}}
            else
            {
                col[v]=col[now]^1;
                q.push(v); 
            }
        }
    }
    return false;
}
int main()
{
    for(int last=0;scanf("%d%d",&n,&m)!=EOF;last=n)
    {
        for(int i=1;i<=last;++i) G[i].clear();
        int ans=0;
        flag=false;
        memset(Map,0,sizeof(Map));
        memset(col,-1,sizeof(col));
        memset(match,0,sizeof(match));
        for(int x,y;m--;)
        {
            scanf("%d%d",&x,&y);
            Map[x][y]=1;
            G[x].push_back(y);
            G[y].push_back(x);  
        }
        for(int i=1;i<=n;++i) if(col[i]==-1) if(rs(i)) break;
        if(flag) {printf("No
");continue;}
        for(int i=1;i<=n;++i)
        {
            memset(vis,0,sizeof(vis));
            ans+=dfs(i);
        }
        printf("%d
",ans);
    }
    return 0;
}