Air Raid （二分图最小路径覆盖）

Consider a town where all the streets are one-way and each street leads from one intersection to another. It is also known that starting from an intersection and walking through town's streets you can never reach the same intersection i.e. the town's streets form no cycles. 

With these assumptions your task is to write a program that finds the minimum number of paratroopers that can descend on the town and visit all the intersections of this town in such a way that more than one paratrooper visits no intersection. Each paratrooper lands at an intersection and can visit other intersections following the town streets. There are no restrictions about the starting intersection for each paratrooper. 

InputYour program should read sets of data. The first line of the input file contains the number of the data sets. Each data set specifies the structure of a town and has the format: 

no_of_intersections 
no_of_streets 
S1 E1 
S2 E2 
...... 
Sno_of_streets Eno_of_streets 

The first line of each data set contains a positive integer no_of_intersections (greater than 0 and less or equal to 120), which is the number of intersections in the town. The second line contains a positive integer no_of_streets, which is the number of streets in the town. The next no_of_streets lines, one for each street in the town, are randomly ordered and represent the town's streets. The line corresponding to street k (k <= no_of_streets) consists of two positive integers, separated by one blank: Sk (1 <= Sk <= no_of_intersections) - the number of the intersection that is the start of the street, and Ek (1 <= Ek <= no_of_intersections) - the number of the intersection that is the end of the street. Intersections are represented by integers from 1 to no_of_intersections. 

There are no blank lines between consecutive sets of data. Input data are correct. 
OutputThe result of the program is on standard output. For each input data set the program prints on a single line, starting from the beginning of the line, one integer: the minimum number of paratroopers required to visit all the intersections in the town. 
Sample Input

2
4
3
3 4
1 3
2 3
3
3
1 3
1 2
2 3

Sample Output

2
1

题目大概：

在一个城镇中，交叉路口和交叉路口之间都有单项道路连接，问最少派多少伞兵，可以把所有交叉路口都走一遍。

思路：

最小路径覆盖问题，在DAG图中寻找尽量少的路径，使得每个节点恰好在一条路径上(不同的路径不可能有公共点)。

刚好和这个题的题意一样。所以，把所有节点i拆为左边点集的i和右边点集的i’，如果DAG图中有i到j的有向边，那么添加一条

二分图的i到j’的无向边。最终DAG的最小路径覆盖数==DAG图的节点数n - 新二分图的最大匹配数m。

代码：

#include 
#include 
#include 
#include 
#include 
#include 
#define MAXN 350
using namespace std;

int G[MAXN][MAXN];
int pipei[MAXN];
bool used[MAXN];
int N,T,M;
void init()
{
    memset(G,0,sizeof(G));

}
void getMap()
{
    int u,v;
    int n=N;
    for(int i=1;i<=M;i++)
    {
        scanf("%d%d",&u,&v);
        G[u][v+n]=G[v+n][u]=1;
    }

}
int find(int x)
{
    for(int i = 1; i <=2*N; i++)
    {
        int y = G[x][i];
        if(y&&!used[i])
        {
            used[i] = true;
            if(pipei[i] == -1 || find(pipei[i]))
            {
                pipei[i] = x;
                return 1;
            }
        }
    }
    return 0;
}

void solve()
{
    int ans = 0;
    memset(pipei, -1, sizeof(pipei));
    for(int i = 1; i <=N; i++)
    {
        memset(used, false, sizeof(used));
        ans += find(i);
    }
    printf("%d\n",N-ans);
}

int main()
{

    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d", &N,&M);
        init();
        getMap();
        solve();
    }
    return 0;
}