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 Kwhich 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
解析: 这题很简单,只需多注意是否是简单环的判断情况即可。
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
const int N = 201,inf = 0x7fffffff;
int edge[N][N];
int main(){
fill(edge[0],edge[0]+N*N,inf);
int n,m,k,num,a,b;
cin>>n>>m;
for(int i=0;i<m;i++){
cin>>a>>b;
edge[a][b] = edge[b][a] = 1;
}
cin>>k;
for(int i=0;i<k;i++){
cin>>num;
vector<int> vec(num),vst(n+1,0);
for(int j=0;j<num;j++)
cin>>vec[j];
if(num != n+1){
cout<<"NO"<<endl;
continue;
}
if(vec[0] != vec[vec.size()-1]){
cout<<"NO"<<endl;
continue;
}
int a,b,flag = 0;
for(int j=0;j<vec.size()-1;j++){
a = vec[j];
b = vec[j+1];
if(edge[a][b] == inf){
flag = 1;
break;
}
vst[a] = 1;
}
vst[b] = 1;
if(flag == 1)
cout<<"NO"<<endl;
else{
int flag1 = 0;
for(int j=1;j<vst.size();j++)
if(vst[j] == 0){
flag1 = 1;
break;
}
if(flag1 == 1)
cout<<"NO"<<endl;
else
cout<<"YES"<<endl;
}
}
return 0;
}