PAT_A1134#Vertex Cover

Source:

PAT A1134 Vertex Cover (25 分)

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:

N​v​​ v[1] v[2]⋯v[N​v​​]

where N​v​​ 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, 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

Keys:

• 图的存储和遍历

Attention:

• vis[n]设为哨兵，少声明一个变量-,-

Code:

 /*
 Data: 2019-05-29 19:57:25
 Problem: PAT_A1134#Vertex Cover
 AC: 19:55

 题目大意：
 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；
 判断所给顶点集合是否为VC集
 输入：
 第一行给出，顶点数N和边数M，均<=1e4
 接下来M行，给出顶点对
 接下来一行，给出测试数K<=100
 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

 基本思路：
 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
 */

 #include<cstdio>
 #include<algorithm>
 const int M=1e4+;
 using namespace std;
 struct node
 {
     int v1,v2;
 }adj[M];
 int vis[M];

 int main()
 {
 #ifdef    ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif

     int n,m,k,v,v1,v2;
     scanf("%d%d", &n,&m);
     for(int i=; i<m; i++)
     {
         scanf("%d%d", &v1,&v2);
         adj[i].v1=v1;
         adj[i].v2=v2;
     }
     scanf("%d", &k);
     while(k--)
     {
         scanf("%d", &v);
         fill(vis,vis+M,);
         for(int i=; i<v; i++){
             scanf("%d", &v1);
             vis[v1]=;
         }
         for(int i=; i<m; i++){
             if(vis[adj[i].v1]== && vis[adj[i].v2]==){
                     vis[n]=;break;
             }
         }
         if(vis[n])  printf("No\n");
         else    printf("Yes\n");
     }

     return ;
 }