A1134. Vertex Cover

A vertex cover of a graph is a set of vertices such that each edge of the graph is incident to at least one vertex of the set. Now given a graph with several vertex sets, you are supposed to tell if each of them is a vertex cover or not.

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 N−1) 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:

N​v​​ v[1] v[2]⋯v[N​v​​]

where N​v​​ is the number of vertices in the set, and v[i]'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

#include<iostream>
#include<cstdio>
using namespace std;
int N, M, edge1[], edge2[], pt = ;
int hashV[] = {};
int main(){
    scanf("%d%d",&N, &M);
    for(int i = ; i < M; i++){
        scanf("%d%d", &edge1[i], &edge2[i]);
    }
    int K;
    scanf("%d", &K);
    for(int i = ; i < K; i++){
        int Nv, tag = ;
        scanf("%d", &Nv);
        fill(hashV, hashV + N, );
        for(int j = ; j < Nv; j++){
            int vv;
            scanf("%d", &vv);
            hashV[vv] = ;
        }
        for(int j = ; j < M; j++){
            if(hashV[edge1[j]] ==  && hashV[edge2[j]] == ){
                tag = ;
                break;
            }
        }
        if(tag == )
            printf("No\n");
        else printf("Yes\n");
    }
    cin >> N;
    return ;
}

总结：

1、题意：给出一个图，然后给出一组点的集合，问这组点能否满足：图中任意一条边均包含至少一个集合中的点。

2、属于模拟题，不需要按照往常存储图的方法（邻接表、邻接矩阵）存储，只需要将每一条边存下来方便遍历即可。可按照edge1[N], edge2[N]的方式存储边。遍历时顺序遍历即可。查看点是否存在，可以将集合中的点都存在哈希表中。