PAT (Advanced Level) 1122. 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

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说实话，这道题虽然题目意思好懂，但是！！有点坑~

首先是在遍历图的时候每遍历完一次都要记得对flag和vis重新初始化一遍~

然后就是我一直第二个测试点过不去，后来又去请教了大神，大神一秒钟指出了我的错误~我在对vector<int> v遍历的时候，只是到t-2就停下来了，然后检查sum是否等于n-1，因为我很想当然的觉得头尾两个数字一样，没必要去检验倒数第二个数字与最后一个数字之间是否联通，第二个测试点应该就是卡这个。。。一定要检查他们是否联通~（好吧。。有时候思考问题的时候就会脑残了，因为只是一味的想着第一个数字的vis已经被置成了true，所以最后一个数字不需要遍历，但是没有考虑到联通的问题啊。。。）

#include<cstdio>#include<queue>#include<cstring>#include<algorithm>#include<vector>using namespace std;const int maxn = 210;int n, m, G[maxn][maxn];int t, flag;bool vis[maxn] = { false };vector<int> v;void bianli() {fill(vis, vis + maxn, 0);flag = 0;int st, ed, sum = 0;st = v[0];ed = v[t - 1];if (st != ed) {return;}for (int i = 0; i < t - 1; i++) {if (vis[v[i]] == false && G[v[i]][v[i + 1]] == 1) {vis[v[i]] = true;sum++;}else {return;}}if (sum == n) {flag = 1;return;}else {return;}}int main() {scanf("%d%d", &n, &m);for (int i = 0; i < m; i++) {int a, b;scanf("%d%d", &a, &b);G[a][b] = G[b][a] = 1;}int s;scanf("%d", &s);for (int i = 0; i < s; i++) {v.clear();scanf("%d", &t);for (int j = 0; j < t; j++) {int temp;scanf("%d", &temp);v.push_back(temp);}bianli();if (flag) {printf("YES\n");}else {printf("NO\n");}}return 0;}