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 ... 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

题目大意：给出一副无向图<N,M>，有K个询问，每个询问包含n个点，判断这n个点组成的序列是否符合哈密顿回路。
         哈密顿回路：包含图中所有节点的一条简单回路（首尾节点相同，且每个顶点只出现一次）。
解题思路：对于每个询问，判断是否符合。出现以下情况就是不符合：
         1、首尾节点不同
         2、询问中的n节点数不等于N+1
         3、路径中没有包含无向图中的所有节点
         4、路径序列中相邻节点之间不连通

#include <cstdio>
#include <vector>
#include <map>
#include <set>
#include <cstdlib>
#include <climits>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define ll long long
const int MAXN = 200 + 5;
const int INF = 0x7f7f7f7f;
template <class XSD> inline XSD f_min(XSD a, XSD b) { if (a > b) a = b; return a; }
template <class XSD> inline XSD f_max(XSD a, XSD b) { if (a < b) a = b; return a; }

bool mp[MAXN][MAXN];
bool judge(int n, int k){
    int x, y;
    set<int>s;
    vector<int>v;
    for(int i=0; i<k; i++){
        scanf("%d", &x);
        s.insert(x);
        v.push_back(x);
    }
    if(s.size()!=n || v[0]!=v[k-1] || k!=(n+1)) return false;
    for(int i=1; i<k; i++) if(!mp[v[i-1]][v[i]]) return false;
    return true;
}
int main(){
    int n, m;
    int x, y;
    while(~scanf("%d%d", &n, &m)){
        memset(mp, false, sizeof mp);
        for(int i=0; i<m; i++){
            scanf("%d%d", &x, &y);
            mp[x][y] = mp[y][x] = true;
        }
        int q, k;
        scanf("%d", &q);
        while(q--){
            scanf("%d", &k);
            if(judge(n, k)) printf("YES\n");
            else printf("NO\n");
        }
    }
    return 0;
}