HDU——2444 The Accomodation of Students

Posted 2020-10-04

　　　　　　　　　　The Accomodation of Students

　　　　　　　　　　　　　　Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
　　　　　　　　　　　　　　　　　　　　Total Submission(s): 7086    Accepted Submission(s): 3167

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：一些学生之间是朋友关系（关系不能传递），现在要将一堆学生分成两堆,使得在同一堆的学生之间没有朋友关系。如果不可以输出“No”，可以的话输出最多可以分出几对小盆友。

思路：

我们先二分图染色，若能被染成两部分的话说明可以被分成两部分，然后再在我们分出的图上面跑最大匹配。若不能被染成两部分直接输出no

代码：

#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define N 510
using namespace std;
bool flag,vis[N];
int n,m,x,y,tot,ans,col[N],girl[N],head[N],map[N][N];
queue<int>q;
struct Edge
{
    int from,to,next;
}edge[N*N];
int read()
{
    int x=0,f=1; char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1; ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘; ch=getchar();}
    return x*f;
}
int add(int x,int y)
{
    tot++;
    edge[tot].to=y;
    edge[tot].next=head[x];
    head[x]=tot;
}
int find(int x)
{
    for(int i=1;i<=n;i++)
    {
        if(!vis[i]&&map[x][i])
        {
            vis[i]=true;
            if(girl[i]==-1||find(girl[i])){girl[i]=x; return 1;}
        }
    }
    return 0;
}
int color(int s)
{
    queue<int>q;
    q.push(s); col[s]=0;
    while(!q.empty())
    {
        int x=q.front();
        for(int i=head[x];i;i=edge[i].next)
        {
            int t=edge[i].to;
            if(col[t]!=-1){if(col[t]==col[x]) {flag=true; return 1;}}
            else
            {
                col[t]=col[x]^1;
                q.push(t);
            }
        }
        q.pop();
    }
    return 0;
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        ans=0;flag=false;tot=0;
        memset(map,0,sizeof(map));
        memset(col,-1,sizeof(col));
        memset(edge,0,sizeof(edge));
        memset(head,0,sizeof(head));
        for(int i=1;i<=m;i++)
        {
            x=read(),y=read();
            map[x][y]=1;
            add(x,y),add(y,x);
        }
        for(int i=1;i<=n;i++)
         if(col[i]==-1)
         {
             if(color(i)) break;
          } 
        if(flag) {printf("No\n"); continue;}
        memset(girl,-1,sizeof(girl));
        for(int i=1;i<=n;i++)
        {
            memset(vis,0,sizeof(vis));
            if(find(i)) ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}