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.
Each input file contains one test case. For each case, the first line gives two positive integers N and M (both no more than 10^4), 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 v[1] v[2]⋯v[Nv]
where Nv is the number of vertices in the set, and v[i]'s are the indices of the vertices.
For each query, print in a line Yes if the set is a vertex cover, or No if not.
题意：给出n个顶点和m条边的信息，再给出k个查询，每个查询中都包含着一些顶点，问图的每一条边是否至少包含其中查询给出的一个顶点，包含的话输出Yes,不包含的话输出No。 思路：如果照着题意这样去的话可能会比较麻烦，不过可以换个思路，在给出的每个查询里面只要所有的顶点包含图中所有的边即可。在输入边的数据的时候，用二维vector存储，在for循环遍历的时候将每个顶点添加每条边的编号，编号从0~n-1,然后在查询的过程中引入hash的vector，将查询设计到的每条边都放入hash中，设值为1，最后只需要遍历所有的边，只要有一个为0的话就输出No,反之输出Yes。
#include<iostream> #include<cstdio> #include<stack> #include<queue> #include<vector> #include<algorithm> #include<cstring> #include<string> #include<cmath> #include<set> #include<map> #include<cctype> #include<cstdlib> #include<ctime> using namespace std; int main() { int n,m,k; scanf("%d%d",&n,&m); vector<int> v[n]; for(int i = 0;i < m;i++) { int t1,t2; scanf("%d%d",&t1,&t2); v[t1].push_back(i); v[t2].push_back(i); } scanf("%d",&k); for(int i = 0;i < k;i++) { int num,nv; scanf("%d",&nv); vector<int> hash(m,0); int flag = 0; for(int j = 0;j < nv;j++) { scanf("%d",&num); for(int l = 0;l < v[num].size();l++) { hash[v[num][l]] = 1; } } for(int j = 0;j < m;j++) { if(hash[j] == 0) { printf("No\n"); flag = 1; break; } } if(flag == 0) { printf("Yes\n"); } } return 0; }