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:

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

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:

?

1 2 3 4 5

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。

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

#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\n");         }else{             printf("Yes\n");         }               }     return 0; }

　　最后还是挺快的，给的600ms，只用了300几ms。