hdoj_1016Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15175    Accepted Submission(s): 6903


Problem DescriptionA 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.

 

Inputn (0 < n < 20). 

OutputThe 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

68

 

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

DFS水題：

#include <iostream>#include <cstring>#include <cmath>using namespace std;int path[30];bool visited[30];int n;bool check(int x){if(x<=1) return false;for(int i=2;i<=sqrt(double(x));i++){if(x % i == 0) return false;}return true;}void DFS(int x, int y){path[y] = x;if(y==n&&check(1+path[n])){for(int i=1;i<n;i++)cout<<path[i]<<" ";cout<<path[n]<<endl;}for(int i=1;i<=n;i++){if(!visited[i]&&check(x+i)){visited[i] = true;DFS(i,y+1);visited[i] = false;}}}int main(){int m = 1;while(cin>>n){memset(visited,false,sizeof(visited));visited[1] = true;printf("Case %d:\n",m++);DFS(1,1);cout<<endl;}return 0;}

ZOJ相同題目，TLE。

看來還得剪枝，會宿舍再想想吧

！！！

當N為奇數的時候，無法構成素數環

int main(){int m = 1;while(scanf("%d",&n)!=EOF){memset(visited,false,sizeof(visited));visited[1] = true;printf("Case %d:\n",m++);if(n%2==1) printf("\n");else{DFS(1,1);printf("\n");}}return 0;}

碉堡了

加幾個函數，就能瞭解深度優先遍曆的基本流程了=。=

#include <iostream>#include <cstring>#include <cmath>#include <cstdio>#include <windows.h>using namespace std;int path[30];bool visited[30];int n;bool check(int x){if(x<=1) return false;for(int i=2;i<=sqrt(double(x));i++){if(x % i == 0) return false;}return true;}void DFS(int x, int y){path[y] = x;if(y==n&&check(1+path[n])){Sleep(1000);for(int i=1;i<n;i++)printf("%d ",path[i]);printf("%d\n",path[n]);}for(int i=2;i<=n;i++){if(!visited[i]&&check(x+i)){visited[i] = true;DFS(i,y+1);visited[i] = false;}}}int main(){int m = 1;while(scanf("%d",&n)!=EOF){memset(visited,false,sizeof(visited));visited[1] = true;printf("Case %d:\n",m++);DFS(1,1);printf("\n");}return 0;}