hdu2444The Accomodation of Students【判断二分图+最大匹配】

我觉得有必要粘一下英文：

The Accomodation of Students

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2502    Accepted Submission(s): 1190

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 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

Source

2008 Asia Harbin Regional Contest Online

 

题意开始读错了，就是一个二分图判断和求最大匹配

 

代码：

#include <iostream>
#include <set>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;

const int maxn = 205;
const int INF = 1000000000;
int n;

int col[maxn];
vector<int> G[maxn];
bool is_bi(int u) {
    for(int i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        if(col[u] == col[v]) return false;
        if(!col[v]) {
            col[v] = 3 - col[u];
            if(!is_bi(v)) return false;
        }
    }
    return true;
}
bool check() {
    for(int i = 1; i <= n; i++) {
        if(G[i].size()) {
            memset(col, 0, sizeof(col));
            col[i] = 1;
            if(!is_bi(i)) return false;
        }
    }
    return true;
}

int vis[maxn];
int Left[maxn];
int Link[maxn];
vector<int> V[maxn];
bool Find(int u) {
    for(int i = 0; i < V[u].size(); i++) {
        int v = V[u][i];
        if(!vis[v]) {
            vis[v] = 1;
            if(Link[v] == -1 || Find(Link[v])) {
                Link[v] = u;
                return true;
            }
        }
    }
    return false;
}
int solve() {
    int cnt = 0;
    memset(Link, -1, sizeof(Link));
    for(int i = 1; i <= n; i++) {
        if(V[i].size()) {
            memset(vis, 0, sizeof(vis));
            if(Find(i)) cnt++;
        }
    }
    return cnt;
}

int main() {
    int m;
    int u, v;
    while(EOF != scanf("%d %d",&n, &m)) {
        for(int i = 0; i <= n; i++) {
            G[i].clear();
            V[i].clear();
        }
        while(m--) {
            scanf("%d %d",&u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
            V[u].push_back(v);
        }
        if(!check()) {
            puts("No");
            continue;
        }
        printf("%d
",solve()) ;
    }
    return 0;
}