PAT 甲级 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 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 <iostream>#include <cstdio>#include <vector>#include <algorithm>using namespace std;int g[10010][10010];int main() {    int n, m;    scanf("%d%d", &n, &m);    for (int i = 0; i < n; i++) {        for (int j = 0; j < n; j++) {            g[i][j] = 0;        }    }    for (int i = 0; i < m; i++) {        int a, b;        scanf("%d%d", &a, &b);        g[a][b] = 1;        g[b][a] = 1;    }    int k;    scanf("%d", &k);    for (int i = 0; i < k; i++) {        int vn;        scanf("%d", &vn);        int flag = 1;        if (vn-1 != n) flag = 0;        vector<int> v(vn);        for (int j = 0; j < vn; j++)            cin >> v[j];        if (v[0] != v[vn - 1]) {            flag = 0;        }        else {            for (int j = 0; j < vn - 1; j++) {                if (!g[v[j]][v[j+1]]) {                    flag = 0;                    break;                }            }            sort(v.begin(), v.end()-1);            for (int j = 0; j < vn - 2; j++) {                if (v[j] == v[j + 1]) {                    flag = 0;                    break;                }            }        }        if (flag)printf("YES\n");        else            printf("NO\n");    }    cin >> n;    return 0;}