pat 1134 Vertex Cover (25分) 超时问题

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 1), 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

这道题目的意思，就是给定一个图，然后K条测试用例，每条测试用例，输入几个点，如果这几个点所连接的边，包含了整个图所有的边，就输出Yes，否则就输出No

这道题目我开始做思路是，是对于每条边，存储它的左端点，存储它的右端点，然后每个测试用例每个点，都去遍历每条边的两个端点，如果左端点或者右端点是这个点，就把这条边加进去，最后看是不是

所有边都加进去了。可惜的是最后两个用例超时了。

于是我换了另外一种存储结构，把每个点连接的边用vector存起来，最后遍历给定的点，看是否包含了所有的边。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<map>
#include<set>
#include<queue>
#include<string>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
int n,m,k,t,child;
vector<int> v[10010]; 


int main(){
#if ONLINE_JUDGE
#else
	freopen("C:\Users\zzloyxt\Desktop\1.txt","r",stdin);	
#endif	
	scanf("%d %d",&n,&m);
	int a,b;
	for(int i=0;i<m;i++){
		scanf("%d %d",&a,&b);
		v[a].push_back(i); //第i号边 
		v[b].push_back(i);
	}
	
	scanf("%d",&k);
	set<int> sets; 
	while(k--){
		sets.clear();
		scanf("%d",&t);
		for(int i=0;i<t;i++){
			scanf("%d",&child);
			//加入其连接的边，set不重复加入 
			for(int j=0;j<v[child].size();j++){
				sets.insert(v[child][j]);
			}
		}
		if(sets.size()!=m){  //没有加入所有边 
			printf("No
");
		}else{
			printf("Yes
");
		}
		
	}
	return 0;
}

　　最后还是挺快的，给的600ms，只用了300几ms。