hdu oj 2444 The Accomodation of Students-二部图

Link:http://acm.hdu.edu.cn/showproblem.php?pid=2444

The Accomodation of Students

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4519    Accepted Submission(s): 2084

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 41 21 31 42 36 51 21 31 42 53 6

 

Sample Output

No3

 

代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <vector>#include <stack>#include <queue>#include <algorithm>using namespace std;int n,m;const int maxn(205);int relation[maxn][maxn];/*用于存储两个认识的人，比如a b互相认识，relation[a][b],relation[b][a]均为1，否则为0*/int color[maxn];//染色法判断是否为二部图int vis[maxn],link[maxn];queue <int > q;bool color_node()//主要是bfs跑整个图{    for(int j=1; j<=n; j++)/*注意可能这个relation表示的图为非连通图        可能一共四个人a,b认识，c,d认识*/    {        if(color[j]==-1)        {            color[j]=0;            q.push(j);            while(!q.empty())            {                int now=q.front();                q.pop();                for(int i=1; i<=n; i++)                {                    if(relation[now][i])                    {                        if(color[i]==-1)                        {                            color[i]=!color[now];                            q.push(i);                        }                        else if(color[i]==color[now])                            return false;                    }                }            }        }    }    return true;}int find(int x)/*匈牙利法判断最大匹配*/{    for(int i=1; i<=n; i++)    {        if(relation[x][i]&&!vis[i])        {            vis[i]=1;            if(!link[i]||find(link[i]))            {                link[i]=x;                return 1;            }        }    }    return 0;}int main(){    int a,b;    int sum;    while(scanf("%d%d",&n,&m)==2)    {        memset(relation,0,sizeof(relation));        memset(color,-1,sizeof(color));        memset(link,0,sizeof(link));        for(int i=0; i<m; i++)        {            scanf("%d%d",&a,&b);            relation[a][b]=relation[b][a]=1;        }        if(!(color_node()))        {            puts("No");            while(!q.empty())                q.pop();        }        else        {            sum=0;            for(int i=1; i<=n; i++)            {                if(!color[i])//仅从一种颜色出发,这样可以降低复杂度，不必sum/2                {                    memset(vis,0,sizeof(vis));                    sum+=find(i);                }            }            cout<<sum<<endl;        }    }    return 0;}

相关连接：http://blog.csdn.net/dark_scope/article/details/8880547

这个链接的文章一个大神写的，感觉很生动