1134. Vertex Cover (25)

青旅半醒 2022-05-28 10:41 110阅读 0赞

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:

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：

代码：

#include<stdio.h>
    #include<map>
    using namespace std;
    struct node
    {
        int x,y;
    }e[10001];
    int main()
    {
        int i,j,n,m,k,t,x,y,v;
        scanf("%d %d",&n,&m);
        for(i=0;i<m;i++)
        {
            scanf("%d %d",&x,&y);
            e[i].x=x;
            e[i].y=y;
        }
        scanf("%d",&k);
        for(i=0;i<k;i++)
        {
            map<int,int> Map;
            scanf("%d",&t);
            for(j=0;j<t;j++)
            {
                scanf("%d",&v);
                Map[v]=1;
            }
            for(j=0;j<m;j++)
            {
                if(Map.find(e[j].x)==Map.end()&&Map.find(e[j].y)==Map.end())
                {
                    break;
                }
            }
            if(j>=m)
            {
                printf("Yes\n");
            }
            else
            {
                printf("No\n");
            }
        }
       return 0;
    }