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:

6 106 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 longconst 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;}