zoukankan      html  css  js  c++  java
  • hdu 1016 DFS

    Prime Ring Problem

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 57319    Accepted Submission(s): 25008


    Problem Description
    A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.

    Note: the number of first circle should always be 1.

     
    Input
    n (0 < n < 20).
     
    Output
    The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order.

    You are to write a program that completes above process.

    Print a blank line after each case.
     

    题意:给你一个1到n个点的环,1必须在第一位,要相邻两点相加为素数。输出所有情况。

    Sample Input
    6 8
     
    Sample Output
    Case 1:
    1 4 3 2 5 6
    1 6 5 2 3 4
     
     
    Case 2:
    1 2 3 8 5 6 7 4
    1 2 5 8 3 4 7 6
    1 4 7 6 5 8 3 2
    1 6 7 4 3 8 5 2
    代码:
    #include<stdio.h>
    #include<string.h>
    int visit[30],a[30],n;//a为第几个数 
    int su(int x)
    {
        for(int i=2;i<x;i++)
            if(x%i==0) return 0;
        return 1;
    }
    void dfs(int x,int y)//x为深度,也是第几位数,y为1~n中的值 
    {
        a[x]=y;
        visit[y]=1;
        if(x==n)//终止条件 
        {
            if(su(a[x]+a[1]))//还要判断一下最后一个与1是否构成素数 
            {
                printf("1");
                for(int i=2;i<=n;i++)
                    printf(" %d",a[i]);
                printf("
    ");
            }
            return ;
        }
        for(int i=1;i<=n;i++)
        {
            if(!visit[i]&&su(a[x]+i))
            {
                dfs(x+1,i);//如果i与a[x]构成素数,搜索下一层 
                visit[i]=0;//回溯 
            }
        }
        return ;
    }
    int main()
    {
        int sum=1;
        while(scanf("%d",&n)!=EOF)
        {
            printf("Case %d:
    ",sum++);
            memset(a,0,sizeof(a));
            dfs(1,1);
            printf("
    ");
        }
        return 0;
    }
  • 相关阅读:
    nginx 配置https 负载均衡
    MyCAT+MySQL搭建高可用企业级数据库集群视频课程
    Java数字签名算法--RSA
    bootstrap在iframe框架中实现由子页面在顶级页面打开模态框(modal)
    bootstrap-treeview 自定义实现双击事件
    Java多线程之内存可见性
    Java实现责任链模式
    JVM(HotSpot) 7种垃圾收集器的特点及使用场景
    jQuery的noConflict以及插件扩展
    JavaScript事件漫谈
  • 原文地址:https://www.cnblogs.com/xiongtao/p/8323652.html
Copyright © 2011-2022 走看看