1122. Hamiltonian Cycle (25)-PAT甲级真题

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 ... 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:

6 106 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

#include <cstdio>int **graph;int isConnected(int *v, int n) {    int pre = v[0];    for (int i = 1; i < n; i++) {        if (graph[pre][v[i]] != 1) {            return 0;        }        pre = v[i];    }    return 1;}int isHamilt(int *v, int n) {    if (v[0] != v[n - 1]) return 0;    int *times = new int[n];    for (int i = 0; i < n; i++) {        times[i] = 0;    }    for (int i = 0; i < n; i++) {        times[v[i]]++;    }        for (int i = 1; i < n; i++) {        if (i == v[0]) {            if (times[i] != 2) {                return 0;            }        } else {            if (times[i] != 1) {                return 0;            }        }    }    return 1;}int main() {    int n = 0, m = 0;    scanf("%d %d", &n, &m);    graph = new int*[n+1];    for (int i = 0; i <= n; i++) {        graph[i] = new int[n+1];    }    for (int i = 0; i <= n; i++) {        for (int j = 0; j <= n; j++) {            graph[i][j] = 0;        }    }    for (int i = 0; i < m; i++) {        int a = 0, b = 0;        scanf("%d %d", &a, &b);        graph[a][b] = graph[b][a] = 1;    }    int k = 0;    scanf("%d", &k);    for (int i = 0; i < k; i++) {        int kn = 0;        scanf("%d", &kn);        int *v = new int[kn];        for (int j = 0; j < kn; j++) {            scanf("%d", &v[j]);        }        if (kn == n + 1 && isConnected(v, kn) && isHamilt(v, kn)) {            printf("YES\n");        } else {            printf("NO\n");        }    }    return 0;}