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 = 20,maxp = 50;
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 = 0;
    is_prime[0] = is_prime[1] = false;
    for(int i =2;i < maxp;i++){
        if(is_prime[i]) prime[cnt++] = i;
        for(int j = 0;j<cnt && prime[j]<=maxp/i;j++){
            is_prime[i*prime[j]] = false;
            if(i%prime[j] == 0) break;
        }
    }
}

void dfs(int cur,int *ans){
    if(cur==n && is_prime[ans[n-1]+ans[0]]){
        for(int i = 0;i < n;i++){
            if(i != 0) printf(" ");
            printf("%d",ans[i]);
        }
        printf("
");
        return;
    }
    for(int i = 2;i <= n;i++){
        if(!vis[i] && is_prime[ans[cur-1]+i]){
            vis[i] = true;
            ans[cur] = i;
            dfs(cur+1,ans);
            vis[i] = false;
        }
    }
}

int main()
{
    //freopen("input.txt","r",stdin);
    int iCase = 1;
    Euler();
    while(~scanf("%d",&n)){
        if(iCase > 1) printf("
");
        memset(vis,false,sizeof(vis));
        ans[0] = 1;
        vis[1] = true;
        printf("Case %d:
",iCase++);
        dfs(1,ans);
    }
    return 0;
}