Air Raid（最小路径覆盖）

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

题目的意思就是：

一个=城镇中有n个路口和m条单项的路径，图是无环图，现在要派一些伞兵去这些成寿寺，要到达所有的路口；

有一些或者没有伞兵可以不去那些路口，只要其他人能完成这么任务；

每个在一个路口着陆了的伞兵可以沿着街去到其他路口；

我们的任务就是求出去执行任务的伞兵最少可以使几个；

 

这个问题是二分图的最小路径覆盖问题；

而路径覆盖的定义就是：

在有向图中找一些路径，使之覆盖了图中的所有顶点，就是任意一个顶点都跟那些路径中的某一条关联，且任何一个顶点

有且只有一个与之关联；一个单独的顶点是一跳路径……最小路径覆盖就是最少的路径覆盖数。

最小路径覆盖=图的顶点数-最大匹配数

关于最小覆盖：http://blog.csdn.net/u014665013/article/details/49870029

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
using namespace std;
#define MAX 125
int n;
int map[MAX][MAX];
bool visited[MAX];
int match[MAX];

bool find(int i)   ///查找当前的i是否可以匹配
{
int j;
for(j=1;j<=n;j++)
{
if(map[i][j]&&!visited[j])
{
visited[j]=1;
if(match[j]==-1||find(match[j]))
{
match[j]=i;
return 1;
}
}
}
return 0;
}
int main()
{
int k,x,y,ans;
int T;
scanf("%d",&T);
while(T--)
{
ans=0;
scanf("%d%d",&n,&k);
memset(map,0,sizeof(map));
memset(match,-1,sizeof(match));

for(int i=0;i<k;i++)//对有意思的进行初始化
{
scanf("%d%d",&x,&y);
map[x][y]=1;
}
for(int i=1;i<=n;i++)
{
memset(visited,0,sizeof(visited));//开始标记为全部没有访问
if(find(i))
ans++;
}
printf("%d\n",n-ans);
}
return 0;
}