PAT 1122 Hamiltonian Cycle

1122 Hamiltonian Cycle

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

总结：这道题目不难，对满足条件的路径输出YES,不满足的输出NO

满足条件：

①：首尾两点相同

②：不能重复访问点，且该访问的点有通路

③：所有的点都要别访问过

代码（不知道哪里错了，扣了一分）：

#include <iostream>
#include <cstring>
using namespace std;

const int N=210;
int g[N][N];
int b[2*N];
int t;
bool st[N];//用于判断当前点是否被访问过

void dfs(){
    if(b[0]!=b[t-1]){//首尾两点是否是相同的点
        puts("NO");
        return;
    }
    int pre=b[0],cur;
    st[0]=true;//因为不会遍历第0个节点，所以这里需要先将它赋值为true,
    //第一个节点会在遍历最后一个节点的时候赋值true
    for(int i=1;i<t;i++){
        cur=b[i];
        if(g[pre][cur]==1 && !st[cur]){//判断是否通路 和该点是否被访问过
            st[cur]=true;
            pre=cur;
        }
        else{
            puts("NO");
            return;
        }
    }
    bool sign=true;
    for(int i=0;i<t;i++)    if(!st[i])  sign=false;//是否全部的点都已经访问过了
    if(sign)    puts("YES");
    else puts("NO");
}
int main(){
    int n,m,k;
    scanf("%d%d",&n,&m);
    
    memset(g,-1,sizeof g);
    for(int i=0;i<m;i++){
        int a,b;
        scanf("%d%d",&a,&b);
        g[a][b]=g[b][a]=1;
    }
    scanf("%d",&k);
    for(int i=0;i<k;i++){
        scanf("%d",&t);
        memset(b,0,sizeof b);
        memset(st,false,sizeof st);
        for(int j=0;j<t;j++)
            scanf("%d",&b[j]);
        dfs();
    }
    return 0;
}

 柳神代码：

满足条件：集合中是否有点重复了，是否是k个点，首尾两点是否相同，路径是否有通路

#include <iostream>
#include <set>
#include <vector>
using namespace std;
int main() {
    int n, m, cnt, k, a[210][210] = {0};
    cin >> n >> m;
    for(int i = 0; i < m; i++) {
        int t1, t2;
        scanf("%d%d", &t1, &t2);
        a[t1][t2] = a[t2][t1] = 1;
    }
    cin >> cnt;
    while(cnt--) {
        cin >> k;
        vector<int> v(k);
        set<int> s;
        int flag1 = 1, flag2 = 1;
        for(int i = 0; i < k; i++) {
            scanf("%d", &v[i]);
            s.insert(v[i]);
        }
        if(s.size() != n || k - 1 != n || v[0] != v[k-1]) flag1 = 0;
        for(int i = 0; i < k - 1; i++)
            if(a[v[i]][v[i+1]] == 0) flag2 = 0;
        printf("%s",flag1 && flag2 ? "YES\n" : "NO\n");
    }
    return 0;
}

好好学习，天天向上！

我要考研！