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 104), 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:
Nv v[1] v[2]⋯v[Nv]
where Nv 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
1 #include <stdio.h> 2 #include <string> 3 #include <iostream> 4 #include <algorithm> 5 #include <vector> 6 #include <string.h> 7 using namespace std; 8 const int maxn=10010; 9 vector<int> adj[maxn]; 10 //int g[maxn][maxn],sav[maxn][maxn]; 11 int vis[maxn]; 12 int n,m,k; 13 int main(){ 14 scanf("%d %d",&n,&m); 15 for(int i=0;i<m;i++){ 16 int c1,c2; 17 scanf("%d %d",&c1,&c2); 18 adj[c1].push_back(c2); 19 adj[c2].push_back(c1); 20 //g[c1][c2]=1; 21 //g[c2][c1]=1; 22 } 23 scanf("%d",&k); 24 while(k--){ 25 int j; 26 scanf("%d",&j); 27 int cnt=0; 28 //memcpy(sav,g,sizeof(g)); 29 fill(vis,vis+maxn,0); 30 for(int i=0;i<j;i++){ 31 int v; 32 scanf("%d",&v); 33 vis[v]=1; 34 for(int q=0;q<adj[v].size();q++){ 35 if(vis[adj[v][q]]==0){ 36 cnt++; 37 } 38 } 39 } 40 if(cnt==m)printf("Yes\n"); 41 else printf("No\n"); 42 } 43 }
注意点：题目读了很久画出来才看懂，就是看给定的点集能不能包含这个图的所有边。
思路就是直接遍历一个点的所有边，把这个点的边条数记录下来，同时记录下这个点，后面有再包含这个边的不能重复计算，遍历完所有点边条数和输入时相等就是yes。
一开始想用二维数组，感觉判断会方便一些，结果又超时又超内存，10的四次方这个级别还是不能用邻接表实现。只有几百的时候可以用邻接表。