HDU 2444 The Accomodation of Students(二分图判定+最大匹配)

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题意：给你N个学生与M对关系,每个关系形如(i,j),表示第i个学生与第j个学生相互认识. 现在要你将N个学生分成两组,每组中的任意两个学生都不相互认识.如果N个学生能分组的话,那么就将这N个学生放到尽量多的双人房中去.每个房间放2个相互认识的人.问你最多需要多少房间?

思路：先判断是否二分图，然后和POJ1466同样的做法

#include<cstdio>#include<cstring>#include<vector>#include<cmath>using namespace std;const int maxn=1000;struct Max_Match{    int n,m;    vector<int> g[maxn];    bool vis[maxn];    int left[maxn];    int color[maxn];    void init(int n)    {        this->n=n;      //  this->m=m;        for(int i=1;i<=n;i++) g[i].clear();        memset(left,-1,sizeof(left));    }    bool bipartite(int u){for (int i = 0;i<g[u].size();i++){int v = g[u][i];if (color[v] == color[u])return false;else if (color[v]==0){color[v]=3-color[u];if (!bipartite(v))return false;}}return true;}    bool match(int u)    {        for(int i=0;i<g[u].size();i++)        {            int v=g[u][i];            if(!vis[v])            {                vis[v]=true;                if(left[v]==-1 || match(left[v]))                {                    left[v]=u;                    return true;                }            }        }        return false;    }        int solve()    {memset(color,0,sizeof(color));for (int i = 1;i<=n;i++)if(!color[i]){color[i]=1;if (!bipartite(i))return -1;}        int ans=0;        for(int i=1;i<=n;i++)        {            memset(vis,0,sizeof(vis));            if(match(i)) ans++;        }        return ans;    }}MM;int T;int main(){int n,m;    while (scanf("%d%d",&n,&m)!=EOF){MM.init(n);for (int i = 1;i<=m;i++){int u,v;scanf("%d%d",&u,&v);MM.g[u].push_back(v);MM.g[v].push_back(u);}int ans = MM.solve();if (ans == -1)printf("No\n");elseprintf("%d\n",ans/2);}}

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

0 0