1122. Hamiltonian Cycle (25)

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:

6 106 23 41 52 53 14 11 66 31 24 567 5 1 4 3 6 2 56 5 1 4 3 6 29 6 2 1 6 3 4 5 2 64 1 2 5 17 6 1 3 4 5 2 67 6 1 2 5 4 3 1

Sample Output:

YESNONONOYESNO用个二维数组记录图，哈密顿回路要求经过各点一次，最后回到原点。#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn = 205;int N, M, K, n;int graph[maxn][maxn], vis[maxn]; int main(){    int x, y;    scanf("%d%d", &N, &M);    for(int i = 0; i < M; i++){        scanf("%d%d", &x, &y);        graph[x][y] = 1;        graph[y][x] = 1;    }    scanf("%d", &K);    int start, tmp, j;    for(int i = 0; i < K; i++){        scanf("%d", &x);        memset(vis, 0, sizeof(vis));        for(j = 0; j < x; j++){            scanf("%d", &y);            if(j){                if(graph[tmp][y] == 1 && vis[y] == 0){                    vis[y] = 1;                    tmp = y;                }else                    break;            } else {                 tmp = start = y;                 vis[start] = 1;            }        }        for(j = j + 1; j < x; j++)            scanf("%d", &y);        if(y != start || graph[tmp][y] != 1)            printf("NO\n");        else {            for(x = 1; x <= N; x++){                if(vis[x] == 0){                    printf("NO\n");                    break;                }            }            if(x == N + 1)                printf("YES\n");        }    }    return 0;}