Air Raid
Time Limit: 1 Sec Memory Limit: 256 MB
题目连接
http://acm.hdu.edu.cn/showproblem.php?pid=1151
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
HINT
题意
在有向图中找一些路径，使之覆盖了图中的所有顶点
题解：
最小路径覆盖=图的顶点数-最大匹配数
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 2001
#define mod 10007
#define eps 1e-9
//const int inf=0x7fffffff; //无限大
const int inf=0x3f3f3f3f;
/*
*/
//**************************************************************************************
inline ll read()
{
int x=0,f=1;char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
return x*f;
}
int ma[maxn][maxn];
int vis[maxn];
int match[maxn];
int n,m;
vector<int> e[maxn];
int dfs(int a)
{
for(int i=0;i<e[a].size();i++)
{
if(vis[e[a][i]]==0)
{
vis[e[a][i]]=1;
if(match[e[a][i]]==-1||dfs(match[e[a][i]]))
{
match[e[a][i]]=a;
return 1;
}
}
}
return 0;
}
int main()
{
int t=read();
while(t--)
{
int n=read();
memset(match,-1,sizeof(match));
for(int i=0;i<=n;i++)
e[i].clear();
int k=read();
for(int i=0;i<k;i++)
{
int x=read(),y=read();
e[x].push_back(y);
}
int ans=0;
for(int i=1;i<=n;i++)
{
memset(vis,0,sizeof(vis));
if(dfs(i)==1)
ans++;
}
printf("%d\n",n-ans);
}
}
