Prime Path
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 17815 | Accepted: 10032 |
Description
— 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
题意：
这个题问你给你两个质数，问你多少次变换能从第一个质数变成第二个质数，每次变换只能变换一个位置上的数。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>
using namespace std;
bool judge_prime(int x) //判断素数
{
if(x == 0 || x == 1)
return false;
else if(x == 2 || x == 3)
return true;
else
{
for(int i = 2; i <= (int)sqrt(x); i++)
if(x % i == 0)
return false;
return true;
}
}
struct qer
{
int x, step;
};
queue <qer> q;
bool vis[233333];
void bfs(int a, int b)
{
q.push((qer){a, 0});
vis[a] = 1;
int x, ans;
while(!q.empty())
{
qer tmp;
tmp = q.front();
q.pop();
x = tmp.x;
ans = tmp.step;
if(x == b)
{
printf("%d\n",ans);
return ;
}
for(int i = 1; i <= 9; i += 2) //个位
{
int s = x / 10 * 10 + i;
if(s != x && !vis[s] && judge_prime(s))
{
vis[s] = 1;
qer temp;
temp.x = s;
temp.step = ans + 1;
q.push(temp);
}
}
for(int i = 0; i <= 9; i++) //十位
{
int s = x / 100 * 100 + i * 10 + x % 10;
if(s != x && !vis[s] && judge_prime(s))
{
vis[s] = 1;
qer temp;
temp.x = s;
temp.step = ans + 1;
q.push(temp);
}
}
for(int i = 0; i <= 9; i++) //百位
{
int s = x / 1000 * 1000 + i * 100 + x % 100;
if(s != x && !vis[s] && judge_prime(s))
{
vis[s] = 1;
qer temp;
temp.x = s;
temp.step = ans + 1;
q.push(temp);
}
}
for(int i = 1; i <= 9; i++) //千位
{
int s = i * 1000 + x % 1000;
if(s != x && !vis[s] && judge_prime(s))
{
vis[s] = 1;
qer temp;
temp.x = s;
temp.step = ans + 1;
q.push(temp);
}
}
}
printf("Impossible\n");
return ;
}
int main()
{
int t;
cin>>t;
while(t--)
{
while(!q.empty())
q.pop();
memset(vis, 0, sizeof(vis));
int a, b;
scanf("%d%d", &a, &b);
bfs(a, b);
}
return 0;
}