Prime Path
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 463 Accepted Submission(s): 305
Problem 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.
1033
1733
3733
3739
3779
8779
8179
The 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.
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
//这题是求从第一个数字变为第二个数字最少须要变几次，每一次仅仅能改变一个数字，而且要满足改变之后的仍然是素数
//从第一个字符到第四个字符相当于迷宫中的四个方向 接着每次改变从0-9注意开头第一个字符不能以0开头
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
bool vis[9999]; //标记数组
char start[10]; //起始数字用字符串处理
int stop; //结尾数字
struct Node
{
char tem[10];
int step;
};
int prime(int n) //推断素数
{
for(int i=2;i*i<=n;i++)
{
if(n%i==0)
return 0;
}
return 1;
}
int bfs(char p[])
{
queue <Node> q;
memset(vis,0,sizeof(vis));
int temp;
Node a;
strcpy(a.tem,p);
a.step=0;
q.push(a);
while(!q.empty())
{
Node b;
b=q.front();
q.pop();
sscanf(b.tem,"%d",&temp); //利用sscanf函数将字符串转换为数字直接与结尾比較
if(temp==stop)
return b.step;
vis[temp]=1;
for(int i=0;i<4;i++)
{
char k;
k=i!=0?'0':'1';
for(;k<='9';k++)
{
Node c;
c=b;
c.tem[i]=k;
sscanf(c.tem,"%d",&temp);
if(prime(temp)&&!vis[temp]) //约束条件
{
c.step++;
q.push(c);
vis[temp]=1;
}
}
}
}
return -1;
}
int main()
{
int t;
scanf("%d",&t);
getchar();
while(t--)
{
scanf("%s%d",start,&stop);
int flag=bfs(start);
int flag=bfs(start);
if(flag==-1)
printf("Impossible\n");
else
printf("%d\n",flag);
}
return 0;
}