POJ1422-Air Raid-二分图-DAG最小路径覆盖

题目地址  http://poj.org/problem?id=1422

二分图基础知识：http://www.cnblogs.com/HITLJR/p/5782110.html

 模板题没有思维含量，但是二分图和DAG最小路径覆盖的关系，需要仔细想明白。

代码

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;
int head[333];
bool check[333];
int matching[333];
struct Edge
{
    int to,next;
}E[333];
int ans,tot;
void addedge(int a,int b)
{
    E[tot].to = b;
    E[tot].next = head[a];
    head[a] = tot++;
}
void init()
{
    memset(E,0,sizeof(E));
    memset(head,-1,sizeof(head));
    memset(matching,-1,sizeof(matching));
    tot = 0;
    ans = 0;
}
bool dfs(int u)
{
    for (int i = head[u] ; i!=-1 ; i = E[i].next)
    {
        int v = E[i].to;
        if (!check[v])
        {
            check[v] = true;
            if (matching[v] == -1 || dfs(matching[v]))
            {
                matching[v] = u;
                matching[u] = v;
                return true;
            }
        }
    }
    return false;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int T;
    cin >> T;
    while (T--)
    {
        init();
        int M,N;
        cin >> M >> N;
        for (int i = 1 ; i <= N ; i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            addedge(a,b+M);
        }
        for (int i = 1; i <= M ; i++)
        {
            if (matching[i] ==-1)
            {
                memset(check,0,sizeof(check));
                if (dfs(i)) ans++;
            }
        }
        cout << M - ans <<endl;
    }

}