标签：25 set No Cover Vertex cin vertex graph each

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​​ [ [

where N​v​​ 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

这道题考察读题，给定顶点，边，给定一个集合，查看每个边是否至少一个点在集合内。我们按题意进行判断即可。

#include <iostream>
#include <unordered_set>
using namespace std;
int N, M, K, tmp, Nv;
pair<int, int> p[99999];
int main() {
    cin >> N >> M;
    for(int i = 0; i < M; i++) 
        cin >> p[i].first >> p[i].second;
    cin >> K;
    while(K--) {
        cin >> Nv;
        unordered_set<int> s;
        while(Nv--) {
            cin >> tmp;
            s.insert(tmp);
        }
        bool flag = true;
        for(int i = 0; i < M; i++) {
            if(s.find(p[i].first) == s.end() && s.find(p[i].second) == s.end()) {
                flag = false;
                break;
            }
        }
        printf("%s\n", flag ? "Yes" : "No");
    }
    return 0;
}

标签：25,set,No,Cover,Vertex,cin,vertex,graph,each
来源： https://www.cnblogs.com/littlepage/p/12827148.html

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