17年春季第三题 PAT甲级 1134 Vertex Cover (25分)

痛定思痛。 2023-07-18 02:59 82阅读 0赞

题目来源:https://pintia.cn/problem-sets/994805342720868352/problems/994805346428633088

备考汇总贴:2020年6月PAT甲级满分必备刷题技巧

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 10^4​​), 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. 10 11
  2. 8 7
  3. 6 8
  4. 4 5
  5. 8 4
  6. 8 1
  7. 1 2
  8. 1 4
  9. 9 8
  10. 9 1
  11. 1 0
  12. 2 4
  13. 5
  14. 4 0 3 8 4
  15. 6 6 1 7 5 4 9
  16. 3 1 8 4
  17. 2 2 8
  18. 7 9 8 7 6 5 4 2

Sample Output:

  1. No
  2. Yes
  3. Yes
  4. No
  5. No

题目大意

给n个结点m条边,再给k个集合。对这k个集合逐个进行判断。每个集合S里面的数字都是结点编号,求问整个图是否符合顶点覆盖的定义(即所有的m条边两端的结点,是否至少一个结点出自集合S中)。如果是,输出Yes否则输出No。

题目分析

用vector和结构体存边,然后遍历每条边,如果存在一条边,它的两个顶点都不在查询的集合里,那就输出No,否则输出Yes。

满分代码

  1. #include<iostream>
  2. #include<vector>
  3. using namespace std;
  4. struct edge{
  5. int a,b;
  6. };
  7. int main(){
  8. int n,m,k,e,x;
  9. scanf("%d %d",&n,&m);
  10. vector<edge> g(m);
  11. for(int i=0;i<m;i++){
  12. scanf("%d %d",&g[i].a,&g[i].b);
  13. }
  14. cin>>k;
  15. while(k--){
  16. cin>>e;
  17. set<int> s;
  18. for(int j=0;j<e;j++){
  19. scanf("%d",&x);
  20. s.insert(x);
  21. }
  22. bool f=true;
  23. for(int i=0;i<m;i++){
  24. if(s.find(g[i].a)==s.end()&&s.find(g[i].b)==s.end()){
  25. cout<<"No"<<endl;
  26. f=false;
  27. break;
  28. }
  29. }
  30. if(f)cout<<"Yes"<<endl;
  31. }
  32. }

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