PAT甲级1122

1122. Hamiltonian Cycle (25)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

The "Hamilton cycle problem" is to find a simple cycle that contains every vertex in a graph. Such a cycle is called a "Hamiltonian cycle".

In this problem, you are supposed to tell if a given cycle is a Hamiltonian cycle.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers N (2< N <= 200), the number of vertices, and M, the number of edges in an undirected graph. Then M lines follow, each describes an edge in the format "Vertex1 Vertex2", where the vertices are numbered from 1 to N. The next line gives a positive integer K which is the number of queries, followed by K lines of queries, each in the format:

n V₁ V₂ ... V_n

where n is the number of vertices in the list, and V_i's are the vertices on a path.

Output Specification:

For each query, print in a line "YES" if the path does form a Hamiltonian cycle, or "NO" if not.

Sample Input:

6 106 23 41 52 53 14 11 66 31 24 567 5 1 4 3 6 2 56 5 1 4 3 6 29 6 2 1 6 3 4 5 2 64 1 2 5 17 6 1 3 4 5 2 67 6 1 2 5 4 3 1

Sample Output:

YESNONONOYESNO

#include<cstdio>#include<set>#include<algorithm>using namespace std;const int maxn = 210;const int INF = 1000000000;int a[maxn];int G[maxn][maxn];int main(){fill(G[0], G[0] + maxn*maxn, INF);int N, M;scanf("%d %d", &N, &M);int v1, v2;for (int i = 0; i < M; i++){scanf("%d %d", &v1, &v2);G[v1][v2] = 1;G[v2][v1] = 1;}int K;scanf("%d", &K);int n;set<int> s; bool flag = false;for (int i = 0; i < K; i++){scanf("%d", &n);for (int j = 0; j < n; j++){scanf("%d", &a[j]);s.insert(a[j]);}if (n != N + 1 || s.size() != N){printf("NO\n");}else{int j = 0;if (n == N + 1 && s.size() == N){for (j = 0; j < n - 1; j++){if (G[a[j]][a[j + 1]] == 1 && a[0] == a[n - 1])continue;else{printf("NO\n");break;}}if (j == n - 1){printf("YES\n");}}}s.clear();}return 0;}