PAT A1122. Hamiltonian Cycle (25)

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

思路分析：

首先要形成一个简单环，且经过所有结点，则意味着整个路径的结点数为图的总结点数加1，只有一个结点出现两次，且出现在队列首尾，其他结点只出现一次。

因此，根据如下步骤判断：

1.输入序列的个数不等于n+1，输出NO

2.输入序列的首尾不同，输出NO

3.输入序列中的数字除首位出现2次外，其余只能出现1次，不符合，输出NO

4.输入序列不连通，输出NO

题解：

#include <cstdio>#include <algorithm>using namespace std;const int MAX = 210;int n, m, k;int g[MAX][MAX] = {0}, query[MAX], counts[MAX] = {0};int main(){scanf("%d %d", &n, &m);while(m--){int u, v;scanf("%d %d", &u, &v);g[u][v] = g[v][u] = 1; }scanf("%d", &k);while(k--){scanf("%d", &m);for(int i = 0; i < m; i++){scanf("%d", &query[i]);}bool judge = true;if(m != n+1 || query[0] != query[m-1]){printf("NO\n");}else{fill(counts, counts+MAX, 0);for(int i = 0; i < m-1; i++){counts[query[i]]++;}for(int i = 1; i <= n; i++){if(counts[i] != 1){judge = false;break;}}if(!judge) printf("NO\n");else{for(int i = 0; i < m-1; i++){if(!g[query[i]][query[i+1]]){judge = false;break;}}if(judge) printf("YES\n");else printf("NO\n");}}}return 0;}