A1134. Vertex Cover
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 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
#include<iostream> #include<cstdio> using namespace std; int N, M, edge1[20001], edge2[20001], pt = 0; int hashV[20001] = {0}; int main(){ scanf("%d%d",&N, &M); for(int i = 0; i < M; i++){ scanf("%d%d", &edge1[i], &edge2[i]); } int K; scanf("%d", &K); for(int i = 0; i < K; i++){ int Nv, tag = 1; scanf("%d", &Nv); fill(hashV, hashV + N, 0); for(int j = 0; j < Nv; j++){ int vv; scanf("%d", &vv); hashV[vv] = 1; } for(int j = 0; j < M; j++){ if(hashV[edge1[j]] == 0 && hashV[edge2[j]] == 0){ tag = 0; break; } } if(tag == 0) printf("No\n"); else printf("Yes\n"); } cin >> N; return 0; }
总结:
1、题意:给出一个图,然后给出一组点的集合,问这组点能否满足:图中任意一条边均包含至少一个集合中的点。
2、属于模拟题,不需要按照往常存储图的方法(邻接表、邻接矩阵)存储,只需要将每一条边存下来方便遍历即可。可按照edge1[N], edge2[N]的方式存储边。遍历时顺序遍历即可。查看点是否存在,可以将集合中的点都存在哈希表中。