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 ... 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

#include<cstdio>
#include<iostream>
#include<vector>
using namespace std;
const int INF = 100000000;
int G[201][201], visit[201] = {0,0};
int main(){
    int N, M;
    fill(G[0], G[0] + 201*201, INF);
    scanf("%d%d", &N, &M);
    for(int i = 0; i < M; i++){
        int v1, v2;
        scanf("%d%d", &v1, &v2);
        G[v1][v2] = G[v2][v1] = 1;
    }
    int K;
    scanf("%d", &K);
    for(int i = 0; i < K; i++){
        int n, s, vi, vj, tag = 1;
        fill(visit, visit + 201, 0);
        scanf("%d%d", &n, &s);
        if(n != N + 1)
            tag = 0;
        visit[s] = 1;
        vi = s;
        for(int j = 1; j < n; j++){
            scanf("%d", &vj);
            if(G[vi][vj] == INF){
                tag = 0;
            }
            visit[vj]++;
            vi = vj;
        }
        if(vj != s)
            tag = 0;
        for(int i = 1; i <= N; i++){
            if(i != s && visit[i] != 1 || i == s && visit[i] != 2)
                tag = 0;
        }
        if(tag == 0)
            printf("NO
");
        else printf("YES
");
    }
    return 0;

}

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。