Air Raid POJ - 1422 【有向无环图（DAG）的最小路径覆盖【最小不相交路径覆盖】 模板题】

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
题意：有n个点和m条有向边，现在要在点上放一些伞兵，然后伞兵沿着图走，直到不能走为止每条边只能是一个伞兵走过，问最少放多少个伞兵。
思路：这是一个很明显的最小路径覆盖，用二分图来做对于这样的一个有向图做最小路径覆盖，首先建图先拆点，将每个点分为两个点，左边是1到n个点，右边是1-n个点然后每一条有向边对应左边的点指向右边的点这样建好图之后求最大匹配数因为最小路径覆盖=点数-最大匹配数

定义：在一个有向图中，找出最少的路径，使得这些路径经过了所有的点。

最小路径覆盖分为最小不相交路径覆盖和最小可相交路径覆盖。

最小不相交路径覆盖：每一条路径经过的顶点各不相同。如图，其最小路径覆盖数为3。即1->3>4，2，5。

最小可相交路径覆盖：每一条路径经过的顶点可以相同。如果其最小路径覆盖数为2。即1->3->4，2->3>5。

特别的，每个点自己也可以称为是路径覆盖，只不过路径的长度是0。

最小不相交路径覆盖的定义：在一个有向图中，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

AC代码：

 #include<vector>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 using namespace std;
 #define maxn 666
 int match[maxn];
 int vis[maxn];
 vector<int> v[maxn];
 int n,m;
 int dfs(int u){
     for(int i=;i<v[u].size();i++){
         int temp=v[u][i];
         if(!vis[temp]){
             vis[temp]=;
             if(match[temp]==||dfs(match[temp])){
                 match[temp]=u;
                 return ;
             }
         }
     }
     return ;
 }
 int main(){
     int _;
     cin>>_;
     while(_--){
         scanf("%d%d",&n,&m);
         for(int i=;i<=n;i++)
             v[i].clear();
         int x,y;
         for(int i=;i<=m;i++){
             scanf("%d%d",&x,&y);
             v[x].push_back(y);
         }
         memset(match,,sizeof(match));
         int ans=;
         for(int i=;i<=n;i++){
             memset(vis,,sizeof(vis));
             if(dfs(i))
                 ans++;
         }
         int res=n-ans;
         printf("%d\n",res);
     }
     return ;
 }