hdu1151 二分图的最小路径覆盖

Air Raid

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3620    Accepted Submission(s): 2380

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
   
   
   
   

 

此题考的是二分图的最小路径覆盖数。

摘自大牛的博客：

有向无环图最小不相交路径覆盖

定义：用最少的不相交路径覆盖所有顶点。

定理：把原图中的每个点V拆成Vx和Vy，如果有一条有向边A->B，那么就加边Ax-By。这样就得到了一个二分图，最小路径覆盖=原图的节点数-新图最大匹配。

简单证明：一开始每个点都独立的为一条路径，总共有n条不相交路径。我们每次在二分图里加一条边就相当于把两条路径合成了一条路径，因为路径之间不能有公共点，所以加的边之间也不能有公共点，这就是匹配的定义。所以有：最小路径覆盖=原图的节点数-新图最大匹配。

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#define Maxn 130
using namespace std;

int adj[Maxn][Maxn];
int match[Maxn];
int vis[Maxn];
int x,y;
int deep;
bool dfs(int u){
    for(int v=1;v<=y;v++){
        if(adj[u][v]&&vis[v]!=deep){
            vis[v]=deep;
            if(match[v]==-1||dfs(match[v])){
                match[v]=u;
                return true;
            }
        }
    }
    return false;
}
int hungary(){
    memset(match,-1,sizeof match);
    memset(vis,-1,sizeof vis);
    int ans=0;
    for(int i=1;i<=x;i++){
        deep=i;
        if(dfs(i)) ans++;
    }
    return ans;
}
int main()
{
    int t,n,m,a,b;
    cin>>t;
    while(t--){
        cin>>n>>m;
        memset(adj,0,sizeof adj);
        for(int i=0;i<m;i++){
            cin>>a>>b;
            adj[a][b]=1;
        }
        x=y=n;
        printf("%d\n",n-hungary());
    }
	return 0;
}