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 104), 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.

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

邻接表,每次记录与某个点有关的边的条数,然后标记这个点。最后判断是否是能有m条边就ok。

代码:

#include <bits/stdc++.h>
using namespace std;
int first[],nex[],u[],v[];
int n,m,k,nv,ve,visited[];
int main()
{
memset(first,-,sizeof(first));
cin>>n>>m;
for(int i = ;i < m;i ++)
{
cin>>u[i]>>v[i];
nex[i] = first[u[i]];
first[u[i]] = i;
u[i + m] = v[i];
v[i + m] = u[i];
nex[i + m] = first[u[i + m]];
first[u[i + m]] = i + m;
}
cin>>k;
while(k --)
{
int c = ;
cin>>nv;
for(int i = ;i < nv;i ++)
{
cin>>ve;
visited[ve] = k;
int ver = first[ve];
while(ver != -)
{
if(visited[v[ver]] != k)
{
c ++;
}
ver = nex[ver];
}
}
if(c == m)cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}
}

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