1122 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 V​1​​ V​2​​ ... 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 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 <bits/stdc++.h>

using namespace std;
int n,v,e,graph[300][300],a,b;
int main(){
    scanf("%d %d",&v,&e);
    for(int i=0;i<e;i++){
        scanf("%d %d",&a,&b);
        graph[a][b]=graph[b][a]=1;
    }
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        int t,last=-1,temp,f[300]={0},bf=1,start=-1;
        scanf("%d",&t);
        
        for(int j=0;j<t;j++){
            scanf("%d",&temp);
            if(!bf)continue;
            if(j==0)start=temp;                     //記錄d開始元素
            if(f[temp]==0)f[temp]=1;
            else if(j!=t-1&&f[temp])bf=0;      //(1)中間數字出現多次
            if(last!=-1&&!graph[temp][last])bf=0;   // (3)如果不聯通，排除
            last=temp;
        }
        if(start!=temp)bf=0;            //（2）檢驗是否成環
        for(int j=1;j<=v;j++)
            if(!f[j]){bf=0;break;}      //（1）檢驗未包含所有點
        if(bf)printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}