hdu 1151 Air Raid（最小路径覆盖）

Air Raid

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4731    Accepted Submission(s): 3170


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)

题意：有一个城镇，它的所有街道都是单行的，并且每条街道都是和两个路口相连。同时已知街道不会形成回路。你的任务是编写程序求最小数量的伞兵，这些伞兵可以访问（visit）所有的路口。对于伞兵的起始降落点不做限制。

思路：最小路径覆盖模板题，直接用匈牙利跑一遍即可。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 150
struct Edge
{
int v,next;
} edge[N*N];
int line
,vis
;
int cnt,head
;
void init()
{
cnt=0;
memset(head,-1,sizeof(head));
memset(line,-1,sizeof(line));
}
void addedge(int u,int v)
{
edge[cnt].v=v;
edge[cnt].next=head[u];
head[u]=cnt++;
}
int can(int t)
{
for(int i=head[t]; i!=-1; i=edge[i].next)
{
int v=edge[i].v;
if(!vis[v])
{
vis[v]=1;
if(line[v]==-1||can(line[v]))
{
line[v]=t;
return 1;
}
}
}
return 0;
}
int main()
{
int T,n,m;
int s,t;
scanf("%d",&T);
while(T--)
{
init();
scanf("%d %d",&n,&m);
for(int i=0; i<m; i++)
{
scanf("%d %d",&s,&t);
addedge(s,t);
}
int ans=0;
for(int i=1; i<=n; i++)
{
memset(vis,0,sizeof(vis));
if(can(i)) ans++;
}
printf("%d\n",n-ans);
}
return 0;
}