PAT甲级——1134 Vertex Cover (25 分)

1134 Vertex Cover (考察散列查找,比较水~)

我先在CSDN上发布的该文章,排版稍好https://blog.csdn.net/weixin_44385565/article/details/88897469 

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N and M (both no more than 1), being the total numbers of vertices and the edges, respectively. Then M lines follow, each describes an edge by giving the indices (from 0 to N1) of the two ends of the edge.

After the graph, a positive integer K (≤ 100) is given, which is the number of queries. Then K lines of queries follow, each in the format:

where Nv​​ is the number of vertices in the set, and ['s are the indices of the vertices.

Output Specification:

For each query, print in a line Yes if the set is a vertex cover, or No if not.

Sample Input:

10 11
8 7
6 8
4 5
8 4
8 1
1 2
1 4
9 8
9 1
1 0
2 4
5
4 0 3 8 4
6 6 1 7 5 4 9
3 1 8 4
2 2 8
7 9 8 7 6 5 4 2

Sample Output:

No
Yes
Yes
No
No

题目大意:对于现有的图,给出K个顶点集合,判断这个集合是否为Vertex Cover;Vertex Cover对于每条边,至少有一个端点处于集合之中。 

思路:题目还是比较水的,用结构数组存储边的两个端点,顶点集合用unordered_set存储,然后遍历图的边判断端点是否在集合之中。。c++的一些stl是真的非常好用,不用unordered_set的话也可以自己写一个哈希表(包括创建、插入和查找函数)。

下面是代码: 

#include<iostream>
#include<unordered_set>
using namespace std;
struct node{
    int v1,v2;//边的两个端点 
};
int main(void)
{
    int N,M,K;
    scanf("%d%d",&N,&M);
    node Edge[M];//用于存储边 
    for(int i=0;i<M;i++)
        scanf("%d%d",&Edge[i].v1,&Edge[i].v2);
    scanf("%d",&K);
    for(int i=0;i<K;i++){
        int Nv,tmp;
        bool flag=true;
        scanf("%d",&Nv);
        unordered_set<int> se;//创建unordered_set集合,底层由哈希表实现 
        for(int j=0;j<Nv;j++){
            scanf("%d",&tmp);
            se.insert(tmp);//将待判定的顶点存入集合 
        }
        /*对于每条边,如果它的每个端点都在集合se中,则为Yes,否则就是No*/
        for(int t=0;t<M;t++){
            if(se.find(Edge[t].v1)==se.end()&&se.find(Edge[t].v2)==se.end()){
                flag=false;
                break;
            }
        }
        if(flag) printf("Yes
");
        else printf("No
");
    }
    return 0;
 } 
原文地址:https://www.cnblogs.com/yinhao-ing/p/10622654.html