HDU

The Accomodation of Students

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

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:有n个学生，其中m对学生相互认识，题目要求你把学生分成两部分，一个部分中的学生相互之间不认识，然后把他们分配到双人间中，房间中的俩人相互认识。不能分成两部分输出“No”(o小写)，否则输出要开多少房间。

思路:模板。

#include <iostream>#include <fstream>#include <cstdio>#include <cstring>#include <queue>#include <stack>#include <vector>#include <map>#include <set>#include <cmath>#include <algorithm>#include <functional>#define inf 0X3f3f3f3fusing namespace std;typedef long long ll;const int MAXN=1e5+10;const int MAX=500+10;const double eps=1e-6;int n,m;vector<int>G[MAX];int vis[MAX];int link[MAX];void init(){    memset(vis,-1,sizeof(vis));    memset(link,-1,sizeof(link));    for(int i=0;i<=n;i++)        G[i].clear();}void addedge(int u,int v){    G[u].push_back(v);}bool dfs(int node,int color){    for(int i=0;i<G[node].size();i++){        int v=G[node][i];        if(vis[v]==-1){            vis[v]=color^1;            if(!dfs(v,vis[v]))                return false;        }        else if(vis[v]==color)            return false;    }    return true;}bool DFS(int node){    for(int i=0;i<G[node].size();i++){        int v=G[node][i];        if(!vis[v]){            vis[v]=1;            if(link[v]==-1||DFS(link[v])){                link[v]=node;                return true;            }        }    }    return false;}int hungary(){    int res=0;    for(int i=1;i<=n;i++){        memset(vis,0,sizeof(vis));        if(DFS(i))            res++;    }    return res;}bool juge(){    for(int i=1;i<=n;i++){        if(G[i].size()&&vis[i]==-1){            vis[i]=0;            if(!dfs(i,0))                return false;        }    }    return true;}int main(){    #ifdef ONLINE_JUDGE    #else    freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    #endif    while(cin>>n>>m){        init();        int u,v;        for(int i=1;i<=m;i++){            scanf("%d%d",&u,&v);            addedge(u,v);            addedge(v,u);        }        if(!juge())            cout<<"No"<<endl;        else            cout<<hungary()/2<<endl;    }    return 0;}