参考题解的 http://user.qzone.qq.com/289065406/blog/1303623014
题目http://poj.org/problem?id=3126
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
大致题意：
给定两个四位素数a b，要求把a变换到b
变换的过程要保证 每次变换出来的数都是一个 四位素数，而且当前这步的变换所得的素数 与 前一步得到的素数 只能有一个位不同，而且每步得到的素数都不能重复。
求从a到b最少需要的变换次数。无法变换则输出Impossible
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
#include<queue>
using namespace std;
using namespace std;
queue<struct node>que;
struct node
{
int prime;
int step;
};
int n,m;
int vis[15000];
int prim(int digit)
{
if(digit==2 || digit==3)
return 1;
else if(digit<=1 || digit%2==0)
return 0;
else if(digit>3)
{
for(int i=3; i*i<=digit; i+=2)
if(digit%i==0)
return 0;
return 1;
}
}
int bfs()
{
int i;
struct node k;
while(!que.empty())
que.pop();
k.prime=n;
k.step=0;
que.push(k);
vis[n]=1;
while(!que.empty())
{
struct node s=que.front();
que.pop();
if(s.prime==m)
{
printf("%d\n",s.step);
return 0;
}
int x=s.prime%10;//获得个位
int y=(s.prime/10)%10;//获得十位
for(i=1; i<=9; i+=2)
{
int a=(s.prime/10)*10+i;
if(a!=s.prime&&!vis[a]&&prim(a))
{
vis[a]=1;
k.prime=a;
k.step=s.step+1;
que.push(k);
}
}
for(i=0; i<=9; i++) //枚举十位
{
int a=(s.prime/100)*100+i*10+x;
if(a!=s.prime && !vis[a] &&prim(a))
{
vis[a]=1;
k.prime=a;
k.step=s.step+1;
que.push(k);
}
}
for(i=0; i<=9; i++) //枚举百位
{
int a=(s.prime/1000)*1000+i*100+y*10+x;
if(a!=s.prime&&vis[a]==0&&prim(a))
{
vis[a]=1;
k.prime=a;
k.step=s.step+1;
que.push(k);
}
}
for(i=1; i<=9; i++) //枚举的千位,保证四位数，千位最少为1
{
int a=s.prime%1000+i*1000;
if(a!=s.prime && !vis[a] && prim(a))
{
vis[a]=1;
k.prime=a;
k.step=s.step+1;
que.push(k);
}
}
}
printf("Impossible\n");
return 0;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d %d",&n,&m);
memset(vis,0,sizeof(vis));
bfs();
}
return 0;
}
g++过了，c++CE
