HDU1035 Prime Ring Problem

文章目錄

• Problem Description
• Input
• Output
• Sample Input
• Sample Output
• Source

                                                   Prime Ring Problem Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 14   Accepted Submission(s) : 10Font: Times New Roman | Verdana | Georgia
Font Size: ← →Problem DescriptionA 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.

Inputn (0 < n < 20).OutputThe 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.Sample Input

68

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

SourceAsia 1996, Shanghai (Mainland China)

解題思路：本題為典型一維深度優先搜尋題。我學深搜時模仿的第一個代碼，不知道怎麼碼解說了。。。。。。

#include<cstdio>#include<cstring>using namespace std;int n;int cil[22];    //已使用資料的儲存int num[22]; //資料使用方式0,1int prime[40]={0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0};   // 素數列表，hass表void dfs(int x){    int i,j;    if(x==n&&prime[cil[0]+cil[x-1]])    //環已滿，頭尾能串連，符合要求    {        printf("1");        for(i=1;i<n;i++)            printf(" %d",cil[i]);        printf("\n");    }    else    {        for(i=2;i<=n;i++)        {            if(!num[i]&&prime[cil[x-1]+i])            {                cil[x]=i;                num[i]=1;                dfs(x+1);                num[i]=0;            }        }    }}int main(){    int i=1;    while(scanf("%d",&n)!=EOF)    {        memset(num,0,sizeof(num));        cil[0]=1;        printf("Case %d:\n",i++);        dfs(1);        printf("\n");    }    return 0;}