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

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#1134 #PAT甲级 #图论 #Vertex Cover #顶点覆盖

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本文介绍了一道PAT甲级编程题，题目要求判断给定的多个顶点集是否为图的顶点覆盖。通过使用vector和结构体存储图的边，并遍历每条边检查其端点是否至少有一个在查询的顶点集中，来实现顶点覆盖的判断。

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:

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

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

#include<iostream>
#include<vector>
using namespace std;
struct edge{
    int a,b;
};
int main(){
    int n,m,k,e,x;
    scanf("%d %d",&n,&m);
    vector<edge> g(m);
    for(int i=0;i<m;i++){
        scanf("%d %d",&g[i].a,&g[i].b);
    }
    cin>>k;
    while(k--){
        cin>>e;
        set<int> s;
        for(int j=0;j<e;j++){
            scanf("%d",&x);
            s.insert(x);
        }
        bool f=true;
        for(int i=0;i<m;i++){
            if(s.find(g[i].a)==s.end()&&s.find(g[i].b)==s.end()){
                cout<<"No"<<endl;
                f=false;
                break;
            }
        }
        if(f)cout<<"Yes"<<endl;
    }
}