1134 Vertex Cover

Posted 2021-02-28 ruruozhenhao

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 [‘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

题意：

　　给出一个图和一个图中部分顶点所构成的集合，判断图中的所有边是不是都和集合中的顶点相关。

思路：

　　利用邻接矩阵来存储图，然后查询与集合中顶点相连的边的条数（注意不要重复）是不是与M（图中边数的总和）相等。

Code：

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 int main() {
 6     int n, m, k;
 7     cin >> n >> m;
 8 
 9     vector<vector<int> > grap(n + 1);
10     int v1, v2;
11     for (int i = 0; i < m; ++i) {
12         cin >> v1 >> v2;
13         grap[v1].push_back(v2);
14         grap[v2].push_back(v1);
15     }
16 
17     cin >> k;
18     int nv, t, edges;
19     unordered_set<int> uset;
20     for (int i = 0; i < k; ++i) {
21         cin >> nv;
22         uset.clear();
23         edges = 0;
24         for (int j = 0; j < nv; ++j) {
25             cin >> t;
26             uset.insert(t);
27             for (int k : grap[t])
28                 if (uset.find(k) == uset.end()) edges++;
29         }
30         if (edges == m)
31             cout << "Yes" << endl;
32         else
33             cout << "No" << endl;
34     }
35     return 0;
36 }