1122. Hamiltonian Cycle (25)

1122. Hamiltonian Cycle (25)
时间限制 300 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序
Standard 作者 CHEN, Yue

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

#define _CRT_SECURE_NO_WARNINGS
#include <algorithm>
#include <iostream>

using namespace std;
const int MaxN = 210;
int N, M;
bool G[MaxN][MaxN];

int main() {
#ifdef _DEBUG
    freopen("data.txt", "r+", stdin);
#endif // _DEBUG
    std::ios::sync_with_stdio(false);

    cin >> N >> M;

    for (int i = 1; i <= N; ++i)G[i][i] = true;
    for (int i = 1; i <= M; ++i) {
        int u, v; cin >> u >> v;
        G[u][v] = G[v][u] = true;
    }

    int Q; cin >> Q;
    for (int i = 0; i < Q; ++i) {
        int n, u, last, oldu; 
        bool passed = true;
        cin >> n >> u; last = oldu = u; 
        int isTra[MaxN] = { 0 }; isTra[last] = true;
        while (--n) {
            cin >> last;
            if (G[oldu][last] == false) {
                passed = false;
            }
            ++isTra[last];
            oldu = last;
        }

        if (!passed || last != u) {
            cout << "NO\n";
            continue;
        }

        int idx = 1;
        while (idx <= N) {
            if (isTra[idx] == 0 || isTra[idx] > 2)break;
            if (isTra[idx] == 2 && idx != u)break;
            ++idx;
        }
        if (idx > N)cout << "YES\n";
        else cout << "NO\n";

    }

    return 0;
}