1122. Hamiltonian Cycle (25)-PAT甲级真题

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 <cstdio>

int **graph;

int isConnected(int *v, int n) {
int pre = v[0];
for (int i = 1; i < n; i++) {
if (graph[pre][v[i]] != 1) {
return 0;
}
pre = v[i];
}
return 1;
}

int isHamilt(int *v, int n) {
if (v[0] != v[n - 1]) return 0;
int *times = new int
;
for (int i = 0; i < n; i++) {
times[i] = 0;
}
for (int i = 0; i < n; i++) {
times[v[i]]++;
}

for (int i = 1; i < n; i++) {
if (i == v[0]) {
if (times[i] != 2) {
return 0;
}
} else {
if (times[i] != 1) {
return 0;
}
}
}
return 1;
}

int main() {
int n = 0, m = 0;
scanf("%d %d", &n, &m);
graph = new int*[n+1];
for (int i = 0; i <= n; i++) {
graph[i] = new int[n+1];
}
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= n; j++) {
graph[i][j] = 0;
}
}
for (int i = 0; i < m; i++) {
int a = 0, b = 0;
scanf("%d %d", &a, &b);
graph[a][b] = graph[b][a] = 1;
}
int k = 0;
scanf("%d", &k);
for (int i = 0; i < k; i++) {
int kn = 0;
scanf("%d", &kn);
int *v = new int[kn];
for (int j = 0; j < kn; j++) {
scanf("%d", &v[j]);
}
if (kn == n + 1 && isConnected(v, kn) && isHamilt(v, kn)) {
printf("YES\n");
} else {
printf("NO\n");
}
}
return 0;
}