UVA524-Prime Ring Problem(搜索剪枝)

Problem UVA524-Prime Ring Problem

Accept：6782  Submit：43814

Time Limit: 3000 mSec

 Problem Description

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 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 ≤ 16)

 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. You are to write a program that completes above process.

 Sample Input

6

8

 

 Sample Ouput

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

题解：回溯法经典题目，回溯就是剪枝嘛，这个题的剪枝就是题意，水题。

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <cstdlib>
 using namespace std;

 const int maxn = ,maxp = ;
 bool vis[maxn];
 int n,ans[maxn];

 bool is_prime[maxp];
 int prime[maxp];

 void Euler(){
     memset(is_prime,true,sizeof(is_prime));
     int cnt = ;
     is_prime[] = is_prime[] = false;
     for(int i =;i < maxp;i++){
         if(is_prime[i]) prime[cnt++] = i;
         for(int j = ;j<cnt && prime[j]<=maxp/i;j++){
             is_prime[i*prime[j]] = false;
             if(i%prime[j] == ) break;
         }
     }
 }

 void dfs(int cur,int *ans){
     if(cur==n && is_prime[ans[n-]+ans[]]){
         for(int i = ;i < n;i++){
             if(i != ) printf(" ");
             printf("%d",ans[i]);
         }
         printf("\n");
         return;
     }
     for(int i = ;i <= n;i++){
         if(!vis[i] && is_prime[ans[cur-]+i]){
             vis[i] = true;
             ans[cur] = i;
             dfs(cur+,ans);
             vis[i] = false;
         }
     }
 }

 int main()
 {
     //freopen("input.txt","r",stdin);
     int iCase = ;
     Euler();
     while(~scanf("%d",&n)){
         if(iCase > ) printf("\n");
         memset(vis,false,sizeof(vis));
         ans[] = ;
         vis[] = true;
         printf("Case %d:\n",iCase++);
         dfs(,ans);
     }
     return ;
 }