1134 Vertex Cover (25 分)

2019-05-09

1134 Vertex Cover (25 分)

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:

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:

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

作者: CHEN, Yue

单位: 浙江大学

时间限制: 600 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

题目比较难懂。给出一个图，然后给出k个集合。要求判断图的每一条边的两个端点是否至少有一个属于对应集合。

分析

存储边，用set存储集合（set的查找速度更快）。然后使用visit数组保存已经查询过的顶点。每次遍历图的所有边，先判断两个端点是否已查询过，都无则在set中查询，更新visit。

代码

#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
using namespace std;
int n,m,k;
int v[10000][10000],visit[10000];
struct edge{
    int a;
    int b;
    edge(){};
    edge(int a,int b):a(a),b(b){};
};
vector<edge> edges;
int main() {
    scanf("%d %d",&n,&m);
    int a,b;
    for(int i=0;i<m;i++){
        scanf("%d %d",&a,&b);
        edge e(a,b);
        edges.push_back(e);
    }
    scanf("%d",&k);
    set<int> s;
    int is;
    for(int i=0;i<k;i++){
        is=1;
        scanf("%d",&a);
        fill(visit,visit+10000,0);
        s.clear();
        for(int j=0;j<a;j++){
            scanf("%d",&b);
            s.insert(b);
        }
        for(int i=0;i<edges.size();i++){
            a=edges[i].a;
            b=edges[i].b;
            if(visit[a]||visit[b])continue;
            if(s.find(a)!=s.end())
                visit[a]=1;
            if(s.find(b)!=s.end())
                visit[b]=1;
            if(visit[a]!=1&&visit[b]!=1)
                is=0;
            if(!is)
                break;
        }
        if(is)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}