HDU 1016 Prime Ring Problem(深搜)
Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32872 Accepted Submission(s): 14544
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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題意:要求1-n之間的所有數組成一個圓環,而且要求相鄰的兩個數相加的和是素數。 思路:因為n的數據范圍只是1-20,所以相加的和最大不會超過40,所以把40以內的素數標記出來。然後DFS深搜,符合題意的直接輸出就可以了。 、
代碼:
#include
#include
#include
#include
#include
#include
using namespace std;
int pv[50];
int n;
int p[50];
int v[50];
void getprime()
{
for(int i=1;i<50;i++)
{
pv[i] = 1;
}
for(int i=1;i<=41;i++)
{
int pi = sqrt(i);
for(int j=2;j<=pi;j++)
{
if(i%j == 0)
{
pv[i] = 0;
break;
}
}
}
}
void DFS(int cnt,int x)
{
if(cnt == n && pv[p[n-1]+p[0]] == 1)
{
for(int i=0;i