1134 Vertex Cover (25 分)
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:
Nv [ [
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
给n个结点m条边,再给k个集合。对这k个集合逐个进行判断。每个集合S里面的数字都是结点编号,
求问整个图所有的m条边两端的结点,是否至少一个结点出自集合S中。如果是,输出Yes否则输出No
直接存是第几条边就行,然后用个set
1 #include <bits/stdc++.h> 2 using namespace std; 3 int n,m,k,p; 4 vector<int> v[10005]; 5 set<int> s; 6 int main(){ 7 cin >> n >> m; 8 int x,y; 9 for(int i = 1; i <= m; i++){ 10 cin >> x >> y; 11 v[x].push_back(i); 12 v[y].push_back(i); 13 } 14 cin >> k; 15 while(k--){ 16 cin >> p; 17 set<int> st; 18 for(int i = 0; i < p; i++){ 19 cin >> x; 20 for(int j = 0; j < v[x].size(); j++){ 21 st.insert(v[x][j]); 22 } 23 } 24 if(st.size() == m){ 25 cout <<"Yes"<<endl; 26 }else{ 27 cout <<"No"<<endl; 28 } 29 } 30 return 0; 31 }