ACM 之 A - Prime Ring 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

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

理解:

学了深搜那么久……在比赛的时候碰到这个题还是一筹莫展。
赛后才慢慢有了方法。用DFS 回溯。

代码部分

#include <cstdio>
using namespace std;
int n,a[21],b[21],t;
int pri(int x){
    for(int i=2;i*i<=x;i++){
        if(x%i==0)
            return 0;
    }
    return 1;
}
void dfs(int num){
    if(n==num&&pri(a[n-1]+1)){
        for(int i=0;i<n;i++)
            printf(i!=n-1?"%d ":"%d\n",a[i]);
    }
    else{
        for(int i=2;i<=n;i++)
            if(!b[i]&&pri(i+a[num-1])){
                a[num]=i;
                b[i]=1;
                dfs(num+1);
                b[i]=0;//这里非常关键，搜完这个数下的情况，要把这个数恢复成未使用的状态才能进行下次的搜索
            }
    }
}
int main(){
    t=1;
    while(scanf("%d",&n)!=EOF){
        a[0]=1;
        for(int i=0;i<n;i++)
            b[i]=0;
        printf("Case %d:\n",t++);
        dfs(1);
        puts("");
    }
}

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