【PAT甲级简单图论】1122 Hamiltonian Cycle (25)

素颜马尾好姑娘i 2023-02-28 04:17 17阅读 0赞

[https://pintia.cn/problem-sets/994805342720868352/problems/994805351814119424][https_pintia.cn_problem-sets_994805342720868352_problems_994805351814119424]

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

### 致谢：[https://blog.csdn.net/liuchuo/article/details/53572051][https_blog.csdn.net_liuchuo_article_details_53572051]（有改动）   题目大意 ###

给出一个图，判断给定的路径是不是哈密尔顿路径（遍历每个顶点的简单回路）。  
简单回路的定义（摘自百度百科）：图的顶点序列中，除了第一个顶点和最后一个顶点相同外，其余顶点不重复出现的回路叫*简单回路*。或者说，若通路或回路不重复地包含相同的边，则它是简单的。

### 分析 ###

1.设置falg1 判断节点是否多走、少走、或走成环  
**2.设置flag2 判断这条路能不能走通（这一点比较坑，如果不写会有个点过不了）**  
3.当falg1、flag2都为1时是哈密尔顿路径，否则不是

#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;
    }

[https_pintia.cn_problem-sets_994805342720868352_problems_994805351814119424]: https://pintia.cn/problem-sets/994805342720868352/problems/994805351814119424
[https_blog.csdn.net_liuchuo_article_details_53572051]: https://blog.csdn.net/liuchuo/article/details/53572051