UVA524-Prime Ring Problem(搜索剪枝)

Posted 2021-01-01 npugen

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。

 

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <cstdlib>
 5 using namespace std;
 6 
 7 const int maxn = 20,maxp = 50;
 8 bool vis[maxn];
 9 int n,ans[maxn];
10 
11 bool is_prime[maxp];
12 int prime[maxp];
13 
14 void Euler(){
15     memset(is_prime,true,sizeof(is_prime));
16     int cnt = 0;
17     is_prime[0] = is_prime[1] = false;
18     for(int i =2;i < maxp;i++){
19         if(is_prime[i]) prime[cnt++] = i;
20         for(int j = 0;j<cnt && prime[j]<=maxp/i;j++){
21             is_prime[i*prime[j]] = false;
22             if(i%prime[j] == 0) break;
23         }
24     }
25 }
26 
27 void dfs(int cur,int *ans){
28     if(cur==n && is_prime[ans[n-1]+ans[0]]){
29         for(int i = 0;i < n;i++){
30             if(i != 0) printf(" ");
31             printf("%d",ans[i]);
32         }
33         printf("
");
34         return;
35     }
36     for(int i = 2;i <= n;i++){
37         if(!vis[i] && is_prime[ans[cur-1]+i]){
38             vis[i] = true;
39             ans[cur] = i;
40             dfs(cur+1,ans);
41             vis[i] = false;
42         }
43     }
44 }
45 
46 int main()
47 {
48     //freopen("input.txt","r",stdin);
49     int iCase = 1;
50     Euler();
51     while(~scanf("%d",&n)){
52         if(iCase > 1) printf("
");
53         memset(vis,false,sizeof(vis));
54         ans[0] = 1;
55         vis[1] = true;
56         printf("Case %d:
",iCase++);
57         dfs(1,ans);
58     }
59     return 0;
60 }