729
HDU1016-Prime Ring Problem
Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26682 Accepted Submission(s): 11908
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)
題意：找出N的各個排列（隻要1開頭的），相鄰兩個數的和必須是素數（包括尾部和頭部）
用來練習廣搜的水題，其實這一題用DFS非常好解決，但是為了練廣搜，我把好好的
一個水題做成了難題......
超級大水題，時間竟然過了
AC代碼：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
struct node
{
int num[25];
int k;
int vis[25];
node(){
k=0;
memset(num,0,sizeof(num));
memset(vis,0,sizeof(vis));
}
};
int vis[25],a[25],prime[110],cnt=1;
int IsPrime(int n)
{
int i;
if(n==1||n==2)
return 1;
for(i=2;i*i<=n;i++)
if(n%i==0) return 0;
return 1;
}
void BFS(int n)
{
queue<node> q;
node a,b;
int i,j,k;
a.num[0]=1;
a.k++;
a.vis[1]=1;
q.push(a);
while(!q.empty())
{
b=q.front();
q.pop();
if(b.k==n)
{
if(prime[b.num[0]+b.num[n-1]]==0)
{
continue;
}
else
{
printf("%d",b.num[0]);//PE了一次
for(j=1;j<n;j++)
printf(" %d",b.num[j]);
puts("");
}
}
for(i=1;i<=n;i++)
{
if(prime[(b.num[b.k-1]+i)]&&b.vis[i]!=1)
{
b.num[b.k++]=i;
b.vis[i]=1;
q.push(b);
b.k--;
b.vis[i]=0;
}
}
}
}
int main()
{
int i,j,n,m;
memset(prime,0,sizeof(prime));
for(i=1;i<=100;i++)//素數打表
{
if(IsPrime(i))
prime[i]=1;
}
while(scanf("%d",&n)!=EOF)
{
memset(vis,0,sizeof(vis));
printf("Case %d:\n",cnt++);
BFS(n);
puts("");
}
return 0;
}
//PS:真他娘折騰人...
最後更新：2017-04-03 05:39:44