https://pintia.cn/problem-sets/994805342720868352/problems/994805346428633088
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 [ [
where Nv is the number of vertices in the set, and ['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
代码:
#include <bits/stdc++.h> using namespace std; const int maxn = 1e5 + 10; int N, M, T, K; int x, vis[maxn]; vector<int> v[maxn]; bool flag = false; int step = 0; int main() { scanf("%d%d", &N, &M); for(int i = 0; i < M; i ++) { int st, en; scanf("%d%d", &st, &en); v[st].push_back(i); v[en].push_back(i); } scanf("%d", &T); while(T --) { step = 0; memset(vis, 0, sizeof(vis)); scanf("%d", &K); for(int i = 0; i < K; i ++){ scanf("%d", &x); for(int j = 0; j < v[x].size(); j ++) { if(vis[v[x][j]] == 0) { vis[v[x][j]] = 1; step ++; } } } if(step == M) printf("Yes "); else printf("No "); } return 0; }
标记边被覆盖的数量 判断数目是否符合 行吧 理解错题意可还行