PAT (Advanced Level) Practice 1134 Vertex Cover （25 分） set

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

 题意：

给出一些顶点，判断是否这些顶点的边能否包含全部的边。

stl中 set的使用。

代码如下：
 

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
#include <map>
#include <set>
using namespace std;
const int maxn=1e4+5;
int n,m,k;
map<int,vector<int> >ma;
int a[1005];
int main()
{
    scanf("%d%d",&n,&m);
    for (int i=0;i<m;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        ma[x].push_back(i);
        ma[y].push_back(i);
    }
    scanf("%d",&k);
    while (k--)
    {
        int l;
        scanf("%d",&l);
        set<int>se;
        for (int i=0;i<l;i++)
        {
            int x;
            scanf("%d",&x);
            for (int j=0;j<ma[x].size();j++)
            {
                se.insert(ma[x][j]);
            }
        }
        if(se.size()==m)
        {
            printf("Yes\n");
        }
        else
        {
            printf("No\n");
        }
    }
    return 0;
}