题目
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
题意
已知图，提供多条路径，求证路径是否为哈密尔顿环（一条包含图中所有顶点的环）
题目分析
路径是哈密尔顿环条件：
- 路径首尾相等
- 路径除了首尾相等，没有其他重复顶点
- 路径顶点数=图顶点数+1（加1是因为首尾相等）
解题思路
Code
#include <iostream>
#include <vector>
#include <set>
using namespace std;
const int maxn=220;
int n,m,e[maxn][maxn];
int main(int argc,char * argv[]) {
scanf("%d %d",&n,&m);
int a=0,b=0;
for(int i=0; i<m; i++) {
scanf("%d %d",&a,&b);
e[a][b]=e[b][a]=1;
}
int k,q;
scanf("%d",&k);
for(int i=0; i<k; i++) {
scanf("%d",&q);
vector<int> v(q);
set<int> s;
bool flag1=true,flag2=true;
for(int j=0; j<q; j++) {
scanf("%d",&v[j]);
s.insert(v[j]);
}
if(s.size()!=n||q-1!=n||v[q-1]!=v[0]) flag1=false;
for(int j=0;j<v.size()-1;j++){
if(e[v[j]][v[j+1]]==0)flag2=false;
}
printf("%s",flag1&&flag2?"YES
":"NO
");
}
return 0;
}