HDU 1016 Prime Ring Problem (回溯法)

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34846    Accepted Submission(s): 15441

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

题意:在1到n构成一个圆环两两相邻的数和为素数，输出所有情况。

思路：感觉搜索最重要的就是找状态，这里的状态就是（当前这个数，已经放置了的个数）。

收获：回溯法：因为有可能再次使用一个数，所以回溯。

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
using namespace std;

const int INF=0x3f3f3f3f;
const double eps=1e-10;
const double PI=acos(-1.0);
#define maxn 500

int n;
int vis[maxn];
int a[maxn];
int judge(int x)
{
    if(x <= 1) return 0;
    int m = floor(sqrt(x) + 0.5);
    for(int i = 2; i <= m; i++)
        if(x%i == 0) return 0;
    return 1;
}
void dfs(int pos, int num)
{
    a[num] = pos;
    if(num == n && judge(pos+1))
    {
        for(int j = 1; j <= n-1; j++)
            printf("%d ", a[j]);
        printf("%d\n", a[n]);
        return;
    }
    for(int i = 2; i <= n; i++)
    {
        int sum  = pos + i;
        if(judge(sum) && !vis[i])
        {
            vis[i] = 1;
            dfs(i, num+1);
            vis[i] = 0;
        }
    }
}
int main()
{
    int cas = 1;
    while(~scanf("%d", &n))
    {
        flag = 1;
        memset(vis, 0, sizeof vis);
        printf("Case %d:\n", cas++);
        dfs(1, 1);
        puts("");
    }
    return 0;
}