A1122. 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 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
#include<cstdio>
#include<iostream>
#include<vector>
using namespace std;
const int INF = ;
int G[][], visit[] = {,};
int main(){
int N, M;
fill(G[], G[] + *, INF);
scanf("%d%d", &N, &M);
for(int i = ; i < M; i++){
int v1, v2;
scanf("%d%d", &v1, &v2);
G[v1][v2] = G[v2][v1] = ;
}
int K;
scanf("%d", &K);
for(int i = ; i < K; i++){
int n, s, vi, vj, tag = ;
fill(visit, visit + , );
scanf("%d%d", &n, &s);
if(n != N + )
tag = ;
visit[s] = ;
vi = s;
for(int j = ; j < n; j++){
scanf("%d", &vj);
if(G[vi][vj] == INF){
tag = ;
}
visit[vj]++;
vi = vj;
}
if(vj != s)
tag = ;
for(int i = ; i <= N; i++){
if(i != s && visit[i] != || i == s && visit[i] != )
tag = ;
}
if(tag == )
printf("NO\n");
else printf("YES\n");
}
return ; }
总结:
1、哈密顿回路:图中所有顶点必须都出现,除了首尾是重复出现外,其它节点仅出现一次。 顶点组成的序列必须是联通的。
2、注意,边读入边做处理时,不要使用break,造成读入数据错乱。
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