hdu 1151 最小路径覆盖

最小路径覆盖是选一个边的集合，使得这些边能够覆盖所有的顶点，并且这些边的数目是最小的。最小路径覆盖=顶点数-最大匹配数

Air Raid

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

Total Submission(s): 3836 Accepted Submission(s): 2531



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

2
1

[align=left]Source[/align]
Asia 2002, Dhaka (Bengal)

//hdu1151 最小路径覆盖

#include<map>
#include<vector>
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<stack>
#include<queue>
#include<set>
#define inf 0x3f3f3f3f
#define mem(a,x) memset(a,x,sizeof(a))

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;

inline int in()
{
int res=0;char c;
while((c=getchar())<'0' || c>'9');
while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
return res;
}

vector<int> v[333];
int vis[333];
int flag;
int match[333];

bool dfs(int x)
{
for(int i=0;i<v[x].size();i++)
{
int t=v[x][i];
if(vis[t]!=flag)
{
vis[t]=flag;
if(match[t]==-1 || dfs(match[t]))
{
match[t]=x;
return 1;
}
}
}
return 0;
}
int main()
{
int t=in();
while(t--)
{
int p=in(),n=in();//p是顶点数目，n是边的个数
for(int i=0;i<=p;i++)v[i].clear();
mem(vis,0);
mem(match,-1);
for(int i=1;i<=n;i++)
{
int tmp=in(),tmp2=in();
v[tmp].push_back(tmp2);
}
int ans=0;
for(int i=1;i<=p;i++)
{
flag=i;
if(dfs(i))ans++;
}
printf("%d\n",p-ans);//顶点数-最大匹配数
}
return 0;
}