zoukankan      html  css  js  c++  java
  • 1122 Hamiltonian Cycle (25分)

    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), 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:

    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

    哈密顿回路问题是一个经典的NP问题,但是这道题上做了简化。我们仅需要判断一个路径是否为哈密顿回路。

    1.第一个输入和最后一个输入一样

    2.路径存在

    3.每个点有且仅经过一次

    #include <iostream>
    #include <set>
    #include <vector>
    #include <map>
    using namespace std;
    int N, M, K, a, b, n;
    set<int> G[99999];
    int main() {
        scanf("%d%d", &N, &M);
        while(M--) {
            scanf("%d%d", &a, &b);
            G[a].insert(b);
            G[b].insert(a);
        }
        scanf("%d", &K);
        while(K--) {
            scanf("%d", &n);
            vector<int> data(n);
            map<int, int> m;
            for(int i = 0; i < n; i++) {
                scanf("%d", &data[i]);
                if(i != 0) m[data[i]]++;
            }
            bool no = false;
            if(data[0] != data[n - 1]) no = true;
            for(int i = 0; i < n - 1; i++) 
                if(G[data[i]].count(data[i + 1]) == 0) no = true;
            if(m.size() != N) no = true;
            for(auto& x: m) if(x.second != 1) no = true;
            printf("%s
    ", no ? "NO": "YES");
        }
        return 0;
    }
  • 相关阅读:
    tomcat9部署到nginx,不能通过nginx访问到tomcat
    解决Linux系统部署webapp,JavaMail 发送邮件javax.mail.MessagingException: 501 Syntax: HELO hostname问题
    先本地仓库中国添加jar包
    IDEA修改pom.xml文件不自动下载的问题
    JavaWeb路径的理解【加不加斜杠又何区别】
    IDEA好用的模板设置
    使用maven启动web项目报错
    MAVEN的学习(图片资源来自黑马程序员)
    option标签不支持单击事件
    immutable.js学习笔记(九)----- Range 与 Repeat
  • 原文地址:https://www.cnblogs.com/littlepage/p/12843664.html
Copyright © 2011-2022 走看看