poj 3126 Prime Path BFS
Prime Path
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 6010 | Accepted: 3440 |
Description
The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices.
— It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
— I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.
Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to minimize the cost. You don't know some very cheap software gurus, do you?
— In fact, I do. You see, there is this programming contest going on... Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.
1033The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.
1733
3733
3739
3779
8779
8179
Input
One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros).
Output
One line for each case, either with a number stating the minimal cost or containing the word Impossible.
Sample Input
3 1033 8179 1373 8017 1033 1033
Sample Output
6 7 0
题意：给两个质数，分别为起点终点。每次可以变动一位为下一个质数，问最少经过多少次变动才能得到终点的质数。
思路：最简单的BFS。
#include<cstdio>
#include<cmath>
#include<cstring>
#include<queue>
using namespace std;
short time[9999];
bool is_prime(int x)
{
if(x%2==0) return false;
for(int i=3;i<=(int)(sqrt((double)x)+0.5);i+=2)
{
if(x%i==0) return false;
}
return true;
}
short bfs(int start,int end)
{
if(start==end) return 0;
int now,tmp,a,b,c,d;
memset(time,-1,sizeof(time));
time[start]=0;
queue<short> que;
que.push(start);
while(!que.empty())
{
now=que.front();que.pop();
d=now;
a=d%10;d/=10;
b=d%10;d/=10;
c=d%10;d/=10;
for(int i=1;i<=9;i++)
{
tmp=i*1000+c*100+b*10+a;
if(tmp==end) return time[now]+1;
if( time[tmp]==-1 && is_prime(tmp) )
{
que.push(tmp);
time[tmp]=time[now]+1;
}
}
for(int i=0;i<=9;i++)
{
tmp=d*1000+i*100+b*10+a;
if(tmp==end) return time[now]+1;
if( time[tmp]==-1 && is_prime(tmp) )
{
que.push(tmp);
time[tmp]=time[now]+1;
}
tmp=d*1000+c*100+i*10+a;
if(tmp==end) return time[now]+1;
if( time[tmp]==-1 && is_prime(tmp) )
{
que.push(tmp);
time[tmp]=time[now]+1;
}
tmp=d*1000+c*100+b*10+i;
if(tmp==end) return time[now]+1;
if( time[tmp]==-1 && is_prime(tmp) )
{
que.push(tmp);
time[tmp]=time[now]+1;
}
}
}
}
int main()
{
int T;scanf("%d",&T);
int start,end;
while(T--)
{
scanf("%d%d",&start,&end);
printf("%d\n",bfs(start,end));
}
return 0;
}