PAT(Advanced Level)1122. Hamiltonian Cycle

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 ... 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

        第一次考pat，结果第三题就卡在这个地方了，哎，最后一个三分怎么也过不了，求哈密顿回路，判断起点终点是否一致，在判断节点数是否是N+1，然后再从起点开始，一个点一个点的走，走一个标记一个，看能否回到起点。最后回来发现我的标记数组重定义了，一个是全局变量，一个是局部变量，改了改就过了，一战没考好，二战终于满分了，终于得到了回报啊

#include <iostream>#include <cstdio>#include <cstring>#include <map>#include <set>using namespace std;int n,m,k,q;int a[205][205],b[205];int main(){    cin>>n>>m;    int u,v;    for(int i = 0;i < m;i++){        cin>>u>>v;        a[u][v] = 1;        a[v][u] = 1;    }    cin>>q;    while(q--){        cin>>k;        set<int>s;        for(int i = 0;i < k;i++){            cin>>b[i];            s.insert(b[i]);        }        if(k!=(n+1)||b[0]!=b[k-1]||s.size()!=n){            printf("NO\n");            continue;        }        int f = 0;        for(int i = 1;i < k;i++){            if(!a[b[i-1]][b[i]]){                f = 1;                printf("NO\n");                break;            }        }        if(!f)printf("YES\n");    }    return 0;}