poj1422(二分图，最小路径覆盖)

Air Raid

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7666		Accepted: 4569

Description

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. 

Input

Your 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. 

Output

The 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

题意：有一个有向图，然后现在有一些伞兵能空降到任何点然后顺着有向图走下去，要使所有伞兵都能遍历图的所有点，问伞兵的最小数量是多少？

显然就是求最小路径覆盖的题了，所以只需要将二分图的最大匹配求出来就ok，然后就是顶点数-最大匹配数，不过注意最小路径的最大匹配是单向的，边不能加双向边。。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
using namespace std;
typedef long long ll;
const int N=555;
bool tu[N][N];
int from[N];///记录右边的点如果配对好了它来自哪里
bool use[N];///记录右边的点是否已经完成了配对
int color[N];
int n,m;
bool dfs(int x)
{
    for(int i=1; i<=m; i++) ///m是右边，所以这里上界是m
        if(!use[i]&&tu[x][i])
        {
            use[i]=1;
            if(from[i]==-1||dfs(from[i]))
            {
                from[i]=x;
                return 1;
            }
        }
    return 0;
}
int hungary()
{
    int tot=0;
    memset(from,-1,sizeof(from));
    for(int i=1; i<=n; i++) ///n是左边，所以这里上界是n
    {
        memset(use,0,sizeof(use));
        if(dfs(i))
            tot++;
    }
    return tot;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int k;
        cin>>n>>k;
        m=n;
        memset(tu,0,sizeof(tu));
        while(k--)
        {
            int a,b;
            cin>>a>>b;
            tu[a][b]=1;
        }
        printf("%d\n",n-hungary());
    }
    return 0;
}