HDU 1151 Air Raid

Air Raid

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5101    Accepted Submission(s): 3439

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

Source

Asia 2002, Dhaka (Bengal)

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3个重要结论：

最小点覆盖数： 最小覆盖要求用最少的点（Ｘ集合或Ｙ集合的都行）让每条边都至少和其中一个点关联。可以证明：最少的点（即覆盖数）＝最大匹配数
最小路径覆盖=最小路径覆盖＝｜N｜－最大匹配数
用尽量少的不相交简单路径覆盖有向无环图Ｇ的所有结点。解决此类问题可以建立一个二分图模型。把所有顶点i拆成两个：Ｘ结点集中的i和Y结点集中的i'，如果有边i->j，则在二分图中引入边i->j'，设二分图最大匹配为m，则结果就是n-m。
二分图最大独立集=顶点数-二分图最大匹配
在Ｎ个点的图G中选出m个点，使这m个点两两之间没有边，求m最大值。
如果图Ｇ满足二分图条件，则可以用二分图匹配来做．最大独立集点数 = N - 最大匹配数。

博客推荐

// 匈牙利算法 的 最小路径覆盖数
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn = 125;
int head[maxn],link[maxn],n,ans,k,Test,tot;
bool vis[maxn];
struct Edge{
    int from,to,next;
}e[maxn*maxn];
void Add_Edge(int u,int v){
    e[++tot].from=u;e[tot].to=v;
    e[tot].next=head[u];head[u]=tot;
}
bool DFS(int u){
    for(int i=head[u];i;i=e[i].next){
        int v=e[i].to;
        if(!vis[v]){
            vis[v]=1;
            if(!link[v] || DFS(link[v])){
                link[v]=u;
                return true;
            }
        }
    }
    return false;
}
void Prepare(){
    memset(e,0,sizeof(e));tot=0;
    memset(head,0,sizeof(head));
    memset(link,0,sizeof(link));
}
int main()
{
    cin>>Test;
    while(Test--){
        Prepare();
        scanf("%d%d",&n,&k);
        for(int i=1,u,v;i<=k;i++){
            scanf("%d%d",&u,&v);
            Add_Edge(u,v);
        }
        int ans=0;
        memset(link,0,sizeof(link));
        for(int i=1;i<=n;i++){
            memset(vis,0,sizeof(vis));
            if(DFS(i))ans++;
        }
        printf("%d
",n-ans);
    }
    return 0;
}