Air Raid

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6562    Accepted Submission(s): 4357

 

Problem 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

Source

Asia 2002, Dhaka (Bengal)

問題連結：HDU1151 Air Raid

問題描述：hdu 1151 最小路徑覆蓋
題目大意：在一個城鎮，有n個路口，m條路，這些路都是單向的，而且路不會形成環，現在要弄一些傘兵去巡查這個城鎮，傘兵只能沿著路的方向走，問最少需要多少傘兵才能把所有的路口搜一遍。
解題思路：求解有向無環圖的最小路徑覆蓋,有向無環圖的最小路徑覆蓋=該圖的頂點數-該圖的最大匹配

AC的C++程式

#include<iostream>
#include<vector>
#include<cstring>

using namespace std;

const int N=200;

vector<int>g[N];
bool vis[N];
int link[N];

bool find(int x)
{
	for(int i=0;i<g[x].size();i++)
	{
		int y=g[x][i];
		if(!vis[y])
		{
			vis[y]=true;
			if(link[y]==-1||find(link[y]))
			{
				link[y]=x;
				return true;
			}
		}
	}
	return false;
}

int main()
{
	int t,n,m,a,b;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		for(int i=0;i<=n;i++)
		  g[i].clear();
		while(m--)
		{
			scanf("%d%d",&a,&b);
			g[a].push_back(b);
		}
		int cnt=0;
		memset(link,-1,sizeof(link));
		for(int i=1;i<=n;i++)
		{
			memset(vis,false,sizeof(vis));
			if(find(i))
			  cnt++;
		}
		printf("%d\n",n-cnt);//最小路徑覆蓋 
	}
	return 0; 
}