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.
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.
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
8
Sample Output
Case 1:
1 4 3 2 5 6
1 6 5 2 3 4
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-;
const double PI=acos(-1.0);
#define maxn 500 int n;
int vis[maxn];
int a[maxn];
int judge(int x)
{
if(x <= ) return ;
int m = floor(sqrt(x) + 0.5);
for(int i = ; i <= m; i++)
if(x%i == ) return ;
return ;
}
void dfs(int pos, int num)
{
a[num] = pos;
if(num == n && judge(pos+))
{
for(int j = ; j <= n-; j++)
printf("%d ", a[j]);
printf("%d\n", a[n]);
return;
}
for(int i = ; i <= n; i++)
{
int sum = pos + i;
if(judge(sum) && !vis[i])
{
vis[i] = ;
dfs(i, num+);
vis[i] = ;
}
}
}
int main()
{
int cas = ;
while(~scanf("%d", &n))
{
flag = ;
memset(vis, , sizeof vis);
printf("Case %d:\n", cas++);
dfs(, );
puts("");
}
return ;
}