hdu 2444 The Accomodation of Students 【二分图匹配】

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.

InputFor 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.

OutputIf 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

tips：给这些点涂上黑白两色，就可以表示分出集合了

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <iomanip>
#define inf 0x3f3f3f3f
typedef long long LL;
using namespace std;

const int maxn = 205;
int n, m;
/*struct edge{
    int from, to;
    edge(int f, int t): from(f), to(t){}
};*/

vector <int> G[maxn];
//vector <edge> edges;
int col[maxn],pre[maxn], flag[maxn];
int check[maxn];

bool dfs(int u, int tar)//黑白标色 找到NO的情况 也就是路径上间隔1的点颜色相同
{
    for(int i = 0; i < G[u].size(); i++){
        int v = G[u][i];
        if(col[v] == -1){
            check[v] = true;
            col[v] = tar ^ 1;
            if(!dfs(v, col[v])){
                return false;
            }
        }
        else if(col[v] == (tar ^ 1)) check[v] = true;
        else if(col[v] == tar) return false;
    }
    return true;
}

int ffind(int u)//找增广路
{
    for(int i = 0; i < G[u].size(); i++){
        int v = G[u][i];
        if(!flag[v]){
            flag[v] = true;
            if(pre[v] == -1 || ffind(pre[v])){
                pre[v] = u;
                return true;
            }
        }
    }
    return false;
}

void init()
{
    for(int i = 0; i < n; i++){
        G[i].clear();
    }
    memset(pre, -1, sizeof(pre));
    memset(col, -1, sizeof(col));
    memset(check, false, sizeof(check));
}
int main()
{
    while(scanf("%d%d",&n,&m) != EOF){
        //int k = 0;
        init();
        for(int i = 0; i < m; i++){
            int f,t;
            scanf("%d%d",&f,&t);
            G[f - 1].push_back(t - 1);
            //edge e(f - 1,t - 1);
            //edges.push_back(e);
        }
        bool tar = false;
        for(int i = 0; i < n; i++){
            if(check[i]) continue;
            col[i] = 1;
            if(!dfs(i, col[i])){
                tar = true;
                break;
            }
        }
        if(tar){
            printf("No\n");
            continue;
        }
        int sum = 0;
        for(int i = 0; i < n; i++){
            memset(flag, false, sizeof(flag));
            sum += ffind(i);
        }
        printf("%d\n",sum);
    }
    return 0;
}