BNUOJ 1541 Air Raid

Air Raid

Time Limit: 1000ms

Memory Limit: 10000KB

This problem will be judged on PKU. Original ID: 1422
64-bit integer IO format: %lld      Java class name: Main

 

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

Dhaka 2002

 

解题：最小路径覆盖。。。

 

路径覆盖是什么？一个PXP的有向图中，路径覆盖就是在图中找一些路径，使之覆盖了图中的所有顶点，且任何一个顶点有且只有一条路径与之关联；（如果把这些路径中的每条路径从它的起始点走到它的终点，那么恰好可以经过图中的每个顶点一次且仅一次）；如果不考虑图中存在回路，那么每条路径就是一个弱连通子集．

 

最小路径覆盖=|P|－最大匹配数

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <climits>
 #include <vector>
 #include <queue>
 #include <cstdlib>
 #include <string>
 #include <set>
 #include <stack>
 #define LL long long
 #define pii pair<int,int>
 #define INF 0x3f3f3f3f
 using namespace std;
 const int maxn = ;
 int mp[maxn][maxn],from[maxn],n,m;
 bool vis[maxn];
 bool dfs(int u){
     for(int v = ; v <= n; v++){
         if(mp[u][v] && !vis[v]){
             vis[v] = true;
             if(from[v] == - || dfs(from[v])){
                 from[v] = u;
                 return true;
             }
         }
     }
     return false;
 }
 int main() {
     int t,u,v,ans,i;
     scanf("%d",&t);
     while(t--){
         scanf("%d %d",&n,&m);
         memset(mp,,sizeof(mp));
         memset(from,-,sizeof(from));
         for(i = ; i < m; i++){
             scanf("%d %d",&u,&v);
             mp[u][v] = ;
         }
         ans = ;
         for(i = ; i <= n; i++){
             memset(vis,false,sizeof(vis));
             if(dfs(i)) ans++;
         }
         printf("%d\n",n-ans);
     }
     return ;
 }