首页 > 代码库 > hdu 1016 Prime Ring Problem(dfs)

hdu 1016 Prime Ring Problem(dfs)

2024-08-07 00:52:27   218人阅读

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28391    Accepted Submission(s): 12644


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.
 

 

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
 
重新做了一遍素数环 没想到还是出了点小问题
 
从第一个数开始dfs 按字典序把符合条件的数计入seq数组
再通过回溯 把所有的情况都遍历并输出
（n==1的情况数据中不会出现 判不判断都可以）
 
#include<cmath>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int n;int vis[50];int seq[50];int prime[100];bool okp(int x){  int i;  for(i=2;i<=sqrt(x);i++)  {    if(x%i==0) break;  }  if(i>sqrt(x)) return true;  return false;}void prim(){  memset(prime,0,sizeof(prime));  int i;  for(i=2;i<=50;i++)  {    if(prime[i]==0)    {      if(okp(i))        prime[i]=1;    }    int k=i,tc=2;    while(k*tc<=50)    {      prime[k*tc]=2;      tc++;    }  }}void dfs(int x){  int i;  if(x>n) return ;  if(x==n&&prime[seq[x]+1]==1)  {    for(i=1;i<=n;i++)    {      printf("%d",seq[i]);      printf("%c",i==n?‘\n‘:‘ ‘);    }  }  else  {    for(i=2;i<=n;i++)    {      if(vis[i]==0&&prime[seq[x]+i]==1)      {        vis[i]=1;        seq[x+1]=i;        dfs(x+1);        vis[i]=0;      }    }  }}int main(){  int cas=1,i,j;  prim();//筛法求素数  while(scanf("%d",&n)!=EOF)  {    printf("Case %d:\n",cas++);    //if(n==1) continue;  //测试数据中不会出现n==1    memset(vis,0,sizeof(vis));    seq[1]=1;    vis[1]=1;        dfs(1);    printf("\n");  }  return 0;}
View Code

 

hdu 1016 Prime Ring Problem(dfs)


联系
我们