Prime Path
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 14374 | Accepted: 8112 |
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
Source
题目大意：
就是给你两个数 a 和 b，让你把 a 变成 b最少需要几步（每次变化 a 与 b 都是素数）
解题思路：
首先判断一下是不是素数，然后枚举个位十位百位和千位，用bfs搜索：
上代码
/* 2015 - 09 - 09 Author: ITAK Motto: 今日的我要超越昨日的我，明日的我要胜过今日的我， 以创作出更好的代码为目标，不断地超越自己。 */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int maxn = 1e4;
const int Max = 15000;
struct node
{
int prime;
int step;
}q[Max];
bool vis[Max];
int a, b;
bool isprime(int data)
{
if(data==2 || data==3)
return 1;
if(data<=1 || data%2==0)
return 0;
else if(data > 3)
for(int i=3; i*i<=data; i+=2)
if(!(data%i))
return 0;
return 1;
}
void bfs()
{
int head, tail;
head = tail = 0;
q[head].prime = a;
q[tail++].step = 0;
vis[a] = 1;
while(head < tail)
{
node x = q[head++];
if(x.prime == b)
{
cout<<x.step<<endl;
return;
}
int numg = x.prime%10;///个位
int nums = (x.prime/10)%10;///十位
///分别枚举个位 十位 百位 千位
for(int i=1; i<=9; i+=2)///个位数是偶数肯定不是素数
{
int dy = (x.prime/10)*10+i;
if(dy!=x.prime && !vis[dy] && isprime(dy))
{
vis[dy] = 1;
q[tail].prime = dy;
q[tail++].step = x.step+1;
}
}
for(int i=0; i<=9; i++)
{
int dy = (x.prime/100)*100+i*10+numg;
if(dy!=x.prime && !vis[dy] && isprime(dy))
{
vis[dy] = 1;
q[tail].prime = dy;
q[tail++].step = x.step+1;
}
}
for(int i=0; i<=9; i++)
{
int dy = (x.prime/1000)*1000+i*100+nums*10+numg;
if(dy!=x.prime && !vis[dy] && isprime(dy))
{
vis[dy] = 1;
q[tail].prime = dy;
q[tail++].step = x.step+1;
}
}
for(int i=1; i<=9; i++)///千位必须大于0
{
int dy = x.prime%1000+1000*i;
if(dy!=x.prime && !vis[dy] && isprime(dy))
{
vis[dy] = 1;
q[tail].prime = dy;
q[tail++].step = x.step+1;
}
}
}
puts("Impossible");
return;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
cin>>a>>b;
memset(vis, 0, sizeof(vis));
bfs();
}
return 0;
}
