HDU 2444 The Accomodation of Students（判断是否为二分图+最大匹配）

The Accomodation of Students

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4084    Accepted Submission(s): 1872

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

 

Source

2008 Asia Harbin Regional Contest Online

 

题意：

        有n个学生，有m对人是认识的，每一对认识的人能分到一间房，问能否把n个学生分成两部分，每部分内的学生互不认识，而两部分之间的学生认识。如果可以分成两部分，就算出房间最多需要多少间，否则就输出No。

分析：

先是要判断是否为二部图，然后求最大匹配。

/*    hdu 2444    题意：判断是否是二分图，并输出最大匹配数     YY：用'临点填色法'判断，相邻点异色，发现同色则不成立        然后匈牙利算法， 求出个数除2     注：匈牙利算法时间复杂度 '邻接表': O(mn)，邻接矩阵: O(n^3)*/  #include <stdio.h>#include <string.h> bool map[210][210];//连接图bool visit[210];//判断是否访问过int link[210];// 当前链接表int judge[210];// 判断二分图时 0-1表 int queue[210];int n,m; bool BFS(){ //二分图BFS判断    int v,start = 0,end = 1;    queue[0] = 1;     for(int i=0;i<=n;i++)         judge[i] = -1;    v = queue[start];    judge[1] = 0;    memset(visit,0,sizeof(visit));     while(start<end)    {        v= queue[start];        for(int i = 1;i <= n; i++)        {            if(map[v][i]){                if(judge[i] == -1){                    judge[i] = (judge[v]+1)%2;                    queue[end++] = i;                   }                else{                    if(judge[i] == judge[v])                        return false;                           }            }        }        start++;    }    return true;} int can(int r){    for(int i=1;i<=n;i++)    {        if(map[r][i] && visit[i] == 0){            visit[i] = 1;            if(link[i]==0 || can(link[i])){                link[i] = r;                return 1;            }        }    }    return 0;} int main(){    while(scanf("%d%d",&n,&m)!=EOF)    {        memset(map,0,sizeof(map));        int a,b;        for(int i=0;i<m;i++)        {            scanf("%d%d",&a,&b);            map[a][b] = 1;            map[b][a] = 1;        }         //judge the bipartite graph         if(!BFS()) {            printf("No\n");continue;        }         //the maximum number of pair        int num = 0;        memset(link,0,sizeof(link));        for(int i=1;i<=n;i++)        {            memset(visit,0,sizeof(visit));                      if(can(i)) num++;        }        printf("%d\n",num/2);    }    return 0;}