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
大水题，40入口的BFS，剪枝后远远没有40入口。
#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;
const int maxp = 10010;
bool isPrime[maxp];
int a, b;
void init()
{
fill(isPrime, isPrime+maxp, true);
for (int i = 2; i < maxp; ++i)
if (isPrime[i])
for (int j = i*i; j < maxp; j += i)
isPrime[j] = false;
}
int bfs()
{
int vis[maxp];
memset(vis, -1, sizeof(vis));
queue<int> q;
q.push(a);
vis[a] = 0;
while (!q.empty())
{
int num = q.front();
q.pop();
if (num == b)
return vis[num];
for (int i = 0; i < 4; ++i)
for (int j = 0; j < 10; ++j)
if (i != 3 || j)
{
int t = (int)(pow(10, i) + 0.5);
int c = num - ((num / t) % 10) * t + j * t;
if (isPrime[c] && vis[c] == -1)
{
q.push(c);
vis[c] = vis[num] + 1;
}
}
}
return -1;
}
int main()
{
init();
int T;
cin >> T;
while (T--)
{
scanf("%d%d", &a, &b);
int ans = bfs();
if (ans == -1)
printf("Impossible\n");
else
printf("%d\n", bfs());
}
return 0;
}
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