1134 Vertex Cover （25 分）图

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 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:

N​v​​ v[1] v[2]⋯v[N​v​​]

where N​v​​ 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
题目大意：
　　图的每条边通过两个顶点来描述，对于图的每条边，如果该边对应的两个顶点中至少有一个在给定顶点集合中，
则输出“Yes”(所有边都要满足)，否则输出“No"。
思路：
　　因为要遍历图的每一条边，所以要使用邻接表存储。对于待判定的顶点集，使用哈希。遍历每一条边，如果这条边
的两个顶点都不在顶点集合中，输出"No"，返回。如果遍历结束还没有返回，则输出”Yes"即可。

#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
#include<string>
#include<map>
#include<set>
#include<stack>
#include<string.h>
#include<cstdio>
#include<cmath>
using namespace std;

vector<vector<int>>graph;

void judge(int temp[],int n)
{
    for(int i=0;i<n;i++)
    {
        for(auto num:graph[i])
        {
            if(temp[i]==0&&temp[num]==0)
            {
                printf("No
");
                return;
            }
        }
    }
    printf("Yes
");
    return;
}


int main()
{
    int n,m;
    cin>>n>>m;
    graph.resize(n);
    for(int i=0;i<m;i++)
    {
        int start,endL;
        cin>>start>>endL;
        graph[start].push_back(endL);
       // graph[endL].push_back(endL);
    }
    cin>>m;
    for(int i=0;i<m;i++)
    {
        int k;
        cin>>k;
        int temp[n];
        memset(temp,0,n*sizeof(int));

        for(int j=0;j<k;j++)
        {
           int a;
           cin>>a;
           temp[a]=1;
        }
        judge(temp,n);
    }

    return 0;
}