PAT_A1134#Vertex Cover
Source:
Description:
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 Mlines 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 ['s are the indices of the vertices.
Output Specification:
For each query, print in a line
Yes
if the set is a vertex cover, orNo
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
Keys:
Attention:
- vis[n]设为哨兵,少声明一个变量-,-
Code:
1 /* 2 Data: 2019-05-29 19:57:25 3 Problem: PAT_A1134#Vertex Cover 4 AC: 19:55 5 6 题目大意: 7 若集合中各顶点的边的集合 = 整个图的边集,称该顶点集为VC集; 8 判断所给顶点集合是否为VC集 9 输入: 10 第一行给出,顶点数N和边数M,均<=1e4 11 接下来M行,给出顶点对 12 接下来一行,给出测试数K<=100 13 接下来K行,首先给出顶点数N,接下来N个数表示N个顶点 14 15 基本思路: 16 遍历各条边,若存在一条边的两个顶点均不在集合中,则非VC集 17 */ 18 19 #include<cstdio> 20 #include<algorithm> 21 const int M=1e4+10; 22 using namespace std; 23 struct node 24 { 25 int v1,v2; 26 }adj[M]; 27 int vis[M]; 28 29 int main() 30 { 31 #ifdef ONLINE_JUDGE 32 #else 33 freopen("Test.txt", "r", stdin); 34 #endif 35 36 int n,m,k,v,v1,v2; 37 scanf("%d%d", &n,&m); 38 for(int i=0; i<m; i++) 39 { 40 scanf("%d%d", &v1,&v2); 41 adj[i].v1=v1; 42 adj[i].v2=v2; 43 } 44 scanf("%d", &k); 45 while(k--) 46 { 47 scanf("%d", &v); 48 fill(vis,vis+M,0); 49 for(int i=0; i<v; i++){ 50 scanf("%d", &v1); 51 vis[v1]=1; 52 } 53 for(int i=0; i<m; i++){ 54 if(vis[adj[i].v1]==0 && vis[adj[i].v2]==0){ 55 vis[n]=1;break; 56 } 57 } 58 if(vis[n]) printf("No\n"); 59 else printf("Yes\n"); 60 } 61 62 return 0; 63 }