PAT 甲级 1122. Hamiltonian Cycle (25)

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 … Vn

where n is the number of vertices in the list, and Vi’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 10

6 2

3 4

1 5

2 5

3 1

4 1

1 6

6 3

1 2

4 5

6

7 5 1 4 3 6 2 5

6 5 1 4 3 6 2

9 6 2 1 6 3 4 5 2 6

4 1 2 5 1

7 6 1 3 4 5 2 6

7 6 1 2 5 4 3 1

Sample Output:

YES

NO

NO

NO

YES

NO

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
int g[10010][10010];

int main() {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
g[i][j] = 0;
}
}
for (int i = 0; i < m; i++) {
int a, b;
scanf("%d%d", &a, &b);
g[a][b] = 1;
g[b][a] = 1;
}
int k;
scanf("%d", &k);
for (int i = 0; i < k; i++) {
int vn;
scanf("%d", &vn);
int flag = 1;
if (vn-1 != n) flag = 0;
vector<int> v(vn);
for (int j = 0; j < vn; j++)
cin >> v[j];
if (v[0] != v[vn - 1]) {
flag = 0;
}
else {
for (int j = 0; j < vn - 1; j++) {
if (!g[v[j]][v[j+1]]) {
flag = 0;
break;
}
}
sort(v.begin(), v.end()-1);
for (int j = 0; j < vn - 2; j++) {
if (v[j] == v[j + 1]) {
flag = 0;
break;
}
}
}
if (flag)printf("YES\n");
else
printf("NO\n");
}
cin >> n;
return 0;
}