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 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.
![](http://acm.hdu.edu.cn/data/images/1016-1.gif)
Note: the number of first circle should always be 1.
![](http://acm.hdu.edu.cn/data/images/1016-1.gif)
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.
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
DFS水题:
#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相同题目,TLE。
看来还得剪枝,会宿舍再想想吧
!!!
当N为奇数的时候,无法构成素数环
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
加几个函数,就能了解深度优先遍历的基本流程了=。=
#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; }
Keep it simple!