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