[HDU 1016]--Prime Ring Problem(回溯)

 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1016

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

 

Source

Asia 1996, Shanghai (Mainland China)

 

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题目大意：有一个整数n，把从1到n的数字无重复的排列成环，且使每相邻两个数（包括首尾）的和都为素数，称为素数环。

　　　　　为了简便起见，我们规定每个素数环都从1开始。有多组测试数据，每组输入一个n(0<n<20)，n=0表示输入结束。

　　　　　输出每组第一行输出对应的Case序号，从1开始。如果存在满足题意叙述的素数环，从小到大输出。

 

解题思路：回溯的思想就是了，一个dfs搞定，最坑爹的是，每输完一组末尾都要加上换行，我没加结果提交wa,明明是pe,各种改,

　　　　　彻底无爱了，Orz~~~

 

代码如下：

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

#define maxn 40
int vis[21], x[21], T, n;
int prime[maxn] = { 1, 1, 0 };
void init()
{
    int i, j;
    for (i = 2; i <= maxn; i++){
        if (!prime[i]){
            for (j = 2; i*j <= maxn; j++)
                prime[i*j] = 1;
        }
    }
}

void dfs(int cur){
    if (cur == n&&!prime[1 + x[n - 1]]){
        for (int i = 0; i < n; i++){
            if (i) cout << ' ';
            cout << x[i];
        }
        cout << endl;
    }
    else for (int i = 2; i <= n; i++){
        if (!vis[i] && !prime[i + x[cur - 1]]){
            x[cur] = i;
            vis[i] = 1;
            dfs(cur + 1);
            vis[i] = 0;
        }
    }
}

int main(){
    init();
    while (cin >> n){
        cout << "Case " << ++T << ':' << endl;
        if (n == 1)
            cout << 1 << endl;
        else if (n & 1)
            cout << endl;
        else{
            memset(vis, 0, sizeof(vis));
            x[0] = 1;
            dfs(1);
        }
        cout << endl;//没加这一句pe来个wa,我也是醉了,各种改,无爱了~~~~
    }
    return 0;
}