1122 Hamiltonian Cycle (25分)

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), 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 V​1​​ V​2​​ ... 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

哈密顿回路问题是一个经典的NP问题，但是这道题上做了简化。我们仅需要判断一个路径是否为哈密顿回路。

1.第一个输入和最后一个输入一样

2.路径存在

3.每个点有且仅经过一次

#include <iostream>
#include <set>
#include <vector>
#include <map>
using namespace std;
int N, M, K, a, b, n;
set<int> G[99999];
int main() {
    scanf("%d%d", &N, &M);
    while(M--) {
        scanf("%d%d", &a, &b);
        G[a].insert(b);
        G[b].insert(a);
    }
    scanf("%d", &K);
    while(K--) {
        scanf("%d", &n);
        vector<int> data(n);
        map<int, int> m;
        for(int i = 0; i < n; i++) {
            scanf("%d", &data[i]);
            if(i != 0) m[data[i]]++;
        }
        bool no = false;
        if(data[0] != data[n - 1]) no = true;
        for(int i = 0; i < n - 1; i++) 
            if(G[data[i]].count(data[i + 1]) == 0) no = true;
        if(m.size() != N) no = true;
        for(auto& x: m) if(x.second != 1) no = true;
        printf("%s\n", no ? "NO": "YES");
    }
    return 0;
}