poj 3126 Prime Path （BFS）

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14276		Accepted: 8045

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.



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

Source

Northwestern Europe 2006
题目大意就是给两个四位的素数a，b，每次替换a的一个数位，让a变到b，但是每次替换中所生成的新四位数也必须是素数。求最少要多少步。
简单的BFS直接替换每个数位即可。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<iostream>
#include<algorithm>
using namespace std;
bool vis[10005];
struct node{
int x,step;
};
node u,v;

bool prime(int u){//judge是否为素数

for(int i=2;i*i<10000;i++)
if(u%i==0)
return false;
return true;

}

int bfs(int start,int end){
u.x=start;
u.step=0;
queue<node>q;
q.push(u);
memset(vis,false,sizeof(vis));
vis[start]=true;
while(!q.empty()){
u=q.front();
q.pop();
if(u.x==end)
return u.step;
for(int i=0;i<=3;i++){
double t1=pow((double)10,i);//pow函数中的10必须转化为double型，若不转化，则会出现以下问题
//当i为2的时候结果为99，当i为4的时候结果为9999，具体原因我也不太清楚，可能是由于
//C++中对pow函数的定义为pow（double，double），如果强加用int可能会有精度损失

int x=(u.x/(int)t1)%10;
for(int j=0;j<=9;j++){
if(j==0&&i==3）//最高位不可以为0continue;
int temp2=u.x+(j-x)*t1;//获得下一个数
if(prime(temp2)&&!vis[temp2]){//如果下一个数不为素数并且没有别访问过
vis[temp2]=true;
v.step=u.step+1;
v.x=temp2;
q.push(v);
}
}

}

}
return -1;
}

int main(){
int  t;
scanf("%d",&t);
while(t--){

int start,end;
scanf("%d%d",&start,&end);
int ans=bfs(start,end);
if(ans==-1)
printf("Impossible\n");
else
printf("%d\n",ans);

}
return 0;
}