7-1 Hamiltonian Cycle
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
Code
#include <stdio.h>
#include <stdlib.h>
int graph[300][300];
void graph_init(){
for(int i=0; i<300; i++){
for(int j=0; j<300; j++)
graph[i][j] = 0;
}
}
int is_seq_ok(int* seq, int size, int total_size){
if(size-1!=total_size || seq[0]!=seq[size-1] ) return 0;
int node[total_size+1];
for(int i=0; i<total_size+1; i++) node[i] = 0;
node[seq[0]] = 1;
for(int i=1; i<size; i++){
if(i<size-1){
if(node[seq[i]]) return 0;
node[seq[i]] = 1;
}
if(graph[seq[i-1]][seq[i]] == 0) {
return 0;
}
}
return 1;
}
int main(int argc, char *argv[]){
graph_init();
int total_size, edge_size;
scanf("%d %d", &total_size, &edge_size);
for(int i=0; i<edge_size; i++){
int n1, n2;
scanf("%d %d", &n1, &n2);
graph[n1][n2] = 1;
graph[n2][n1] = 1;
}
int case_count;
scanf("%d", &case_count);
for(int i=0; i<case_count; i++){
int seq_size;
scanf("%d", &seq_size);
int seq[seq_size];
for(int j=0; j<seq_size; j++){
int num;
scanf("%d", &num);
seq[j] = num;
}
if(is_seq_ok(seq, seq_size, total_size)) printf("YES\n");
else printf("NO\n");
}
}