PAT甲级——1134 Vertex Cover （25 分)

1134 Vertex Cover （考察散列查找，比较水~)

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：对于现有的图，给出K个顶点集合，判断这个集合是否为Vertex Cover；Vertex Cover对于每条边，至少有一个端点处于集合之中。 

思路：题目还是比较水的，用结构数组存储边的两个端点，顶点集合用unordered_set存储，然后遍历图的边判断端点是否在集合之中。。c++的一些stl是真的非常好用，不用unordered_set的话也可以自己写一个哈希表（包括创建、插入和查找函数）。

下面是代码： 

#include<iostream>
#include<unordered_set>
using namespace std;
struct node{
	int v1,v2;//边的两个端点 
};
int main(void)
{
	int N,M,K;
	scanf("%d%d",&N,&M);
	node Edge[M];//用于存储边 
	for(int i=0;i<M;i++)
		scanf("%d%d",&Edge[i].v1,&Edge[i].v2);
	scanf("%d",&K);
	for(int i=0;i<K;i++){
		int Nv,tmp;
		bool flag=true;
		scanf("%d",&Nv);
		unordered_set<int> se;//创建unordered_set集合，底层由哈希表实现 
		for(int j=0;j<Nv;j++){
			scanf("%d",&tmp);
			se.insert(tmp);//将待判定的顶点存入集合 
		}
		/*对于每条边，如果它的每个端点都在集合se中，则为Yes，否则就是No*/
		for(int t=0;t<M;t++){
			if(se.find(Edge[t].v1)==se.end()&&se.find(Edge[t].v2)==se.end()){
				flag=false;
				break;
			}
		}
		if(flag) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
 }