PAT 1122 Hamiltonian Cycle[比较一般]

1122 Hamiltonian Cycle （25 分）

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

题目大意：判断给出的路径是否是哈密顿回路，哈密顿回路是一个简单回路，包含图中的每一个点，

我的AC：

#include <iostream>
#include <vector>
#include<cstdio>
#include <map>
using namespace std;
#define inf 9999

int g[205][205];
int vis[205];
int main() {
    int n,m,f,t;
    cin>>n>>m;
    fill(g[0],g[0]+205*205,inf);
    for(int i=0;i<m;i++){
        cin>>f>>t;
        g[f][t]=1;
        g[t][f]=1;
    }
    int k,ct;
    cin>>k;
    while(k--){
        fill(vis,vis+205,0);
        cin>>ct;
        vector<int> path(ct);
        for(int i=0;i<ct;i++){
            cin>>path[i];
        }
        if(path[0]!=path[ct-1]){//首先需要保证两者是相同的。
            cout<<"NO\n";continue;
        }
        bool flag=false;
        for(int i=0;i<ct-1;i++){
           if(g[path[i]][path[i+1]]==inf){//如果两点之间，没有路径。
                cout<<"NO\n";
                flag=true;
                break;
           }
           if(vis[path[i+1]]==1){//如果重复访问那么就不是简单路径，
                cout<<"NO\n";
                flag=true;break;
           }
           vis[path[i+1]]=1;
          // cout<<path[i+1]<<‘\n‘;
        }
        if(!flag){//这里还需要判断是否是所有的点都已经访问过。
            bool fg=false;
            for(int i=1;i<=n;i++){//这里是从1开始判断啊喂！！！
                if(vis[i]==0){
                    cout<<"NO\n";
                    fg=true;break;
                }
            }
            if(!fg)cout<<"YES\n";
        }
    }
    return 0;
}

//本来很简单的一道题，两个周没打算法代码了，生疏了。

1.点标号是从1开始的所以 最后判断所有的点是否被遍历过，是从1开始循环的，

2.比较简单，就是几个判断情况，使用邻接矩阵存储图，不是邻接表。

原文地址：https://www.cnblogs.com/BlueBlueSea/p/9689706.html