HDU1151：Air Raid（最小边覆盖）

Posted 2021-01-18 heyuhhh

Air Raid

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6528    Accepted Submission(s): 4330

题目链接：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:

2
1

题意：

最少需要多少个跳伞兵，可以到达所有的点，如果点已经被别的伞兵到达过，则不再被重复到达。

 

题解：

最小边覆盖模板题。

最小边覆盖指的就是最少多少边可以把所有点相连接（每个点一条边与之相连），求法就是连边时构造一个x‘集合，将x集合与x‘集合中的元素相连。最终的答案就是点的总数减去最大匹配数。

证明方法可以参考《训练指南》，大致意思就是:最小边覆盖<=>最少结尾点<=>最多非结尾点<=>二分图最大匹配。这里的结尾点指的就是简单匹配中最后的那个结点。

这题还有个进阶版：http://poj.org/problem?id=2594

 

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#define mem(x) memset(x,0,sizeof(x))
using namespace std;

const int N = 510 ;
int n,t,m,ans;
int check[N],match[N],map[N][N];

inline int dfs(int x){
    for(int i=n+1;i<=2*n;i++){
        if(map[x][i] && !check[i]){
            check[i]=1;
            if(!match[i] || dfs(match[i])){
                match[i]=x;
                return 1;
            }
        }
    }
    return 0;
}

int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        mem(map);mem(match);ans=0;
        for(int i=1,x,y;i<=m;i++){
            scanf("%d%d",&x,&y);
            map[x][y+n]=1;
        }
        for(int i=1;i<=n;i++){
            mem(check);
            if(dfs(i)) ans++;
        }
        printf("%d
",n-ans);
    }
    return 0;
}