hdu1016(dfs实现环状全排序)
2016-03-25 22:11
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Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 39348 Accepted Submission(s): 17338
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(顺时针方向的) and anticlockwisely(逆时针的). The order of numbers must satisfy the above requirements. Print solutions in lexicographical(辞典编纂的) 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
题解:第一个数从2到n开始往下找,标记此数,然后继续第二个数从2到n开始往下找,标记此数,找到了第n个数的时候把它打印出来,然后往回退一个,继续往下找,以此类推,直到找完所有的情况为止。
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 39348 Accepted Submission(s): 17338
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(顺时针方向的) and anticlockwisely(逆时针的). The order of numbers must satisfy the above requirements. Print solutions in lexicographical(辞典编纂的) 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
题解:第一个数从2到n开始往下找,标记此数,然后继续第二个数从2到n开始往下找,标记此数,找到了第n个数的时候把它打印出来,然后往回退一个,继续往下找,以此类推,直到找完所有的情况为止。
#include<iostream> #include<cstdio> #include<cmath> #include<cstring> using namespace std; //int prime[42]={0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1}; int vis[21],a[21],n; bool prime(int n){ int k,i; if(n<2)return false; for(i=2;i<=(int)sqrt(n);i++){ if(n%i==0)return false; } return true; } void dfs(int k){ int i; if(k==n&&prime(a[0]+a[n-1])){ for(i=0;i<n-1;i++) printf("%d ",a[i]); printf("%d\n",a[n-1]); } else{ for(i=2;i<=n;i++){ if(!vis[i]){ if(prime(a[k-1]+i)){ vis[i]=1; a[k++]=i; dfs(k); vis[i]=0; k--; } } } } } int main(){ int i; i=1; while(scanf("%d",&n)!=EOF){ printf("Case %d:\n",i); memset(vis,0,sizeof(vis)); a[0]=1; dfs(1); i++; printf("\n"); } return 0; }
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