HDU 1016 Prime Ring Problem 深搜



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

6
8 

Sample Output

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 

题意：就是给定一个数n，让你找出所有组成的圆圈,并且组成的圆圈相邻的两个数的和必须是质数，圆圈的首位是1，

思路：每次从2开始搜索，并且用过的数标记掉

AC代码：

#include <iostream>
#include <cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
int n,sc=1;
int a[21];
int visit[21]={0};
int ok(int x,int y){//判断质数
    for(int i=2;i<=sqrt(x+y);++i){
        if((x+y)%i==0){
            return 0;
        }
    }
    return 1;
}
void dfs(int x){//x表示已经选的数的个数
    if(x==n&&ok(a[x],1)){
            printf("%d",a[1]);
        for(int j=2;j<=n;j++){
            printf(" %d",a[j]);
        }
        printf("\n");
    return;
    }
    for(int i=2;i<n+1;i++){
        if(visit[i]==0&&ok(a[x],i)){
            a[++x]=i;
            visit[i]=1;
            dfs(x);
            visit[i]=0;
            x--;
        }
    }
}
int main(){
    while(~scanf("%d",&n)){
           a[0]=0;
           a[1]=1;
           printf("Case %d:\n",sc++);
           dfs(1);
           printf("\n");
    }
   return 0;
}

注意：在输入的时候！=EOF，求质数的过程中<=sqrt();

这一题和HDU 1258 都用到了先把一些值储存再打印。。。做过那道这道就简单了。。。。。。。。。。fighting。。。

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