[kuangbin带你飞]专题一 简单搜索-F - Prime Path

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15921		Accepted: 8987

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

分析：一个数字枚举出所有的状态，就是bfs嘛。水题，双向bfs如何实现是个问题。

code：

#include <cstdio>
#include <cstring>
#include <queue>
#include <iostream>

using namespace std;
typedef long long ll;
typedef struct{
int number[4];
int wal;
}data;
data star,en;
int n,m;
int is_prim[10005],vis[10005];
int dir[]={1,10,100,1000};

void init()
{
cin>>n>>m;
memset(vis,0,sizeof vis);
star.number[0]=n/1000;
star.number[1]=n/100%10;
star.number[2]=n/10%10;
star.number[3]=n%10;
en.number[0]=m/1000;
en.number[1]=m/100%10;
en.number[2]=m/10%10;
en.number[3]=m%10;
star.wal=0;
en.wal=0;
}

void prim_()
{
memset(is_prim,0,sizeof is_prim);
for(int i=2;i<=9999;i++)
{
if(is_prim[i]==1)
continue;
for(int j=i*2;j<=9999;j+=i)
{
is_prim[j]=1;
}
}

}

void bfs()
{
queue <data> qf;
queue <data> qb;
vis
=1;

qf.push(star);
qb.push(en);

while(qf.size())
{
data nos,noe;
nos=qf.front();
noe=qb.front();
int stemp=0,etemp=0;
for(int i=0;i<4;i++)
{
stemp=stemp*10+nos.number[i];
etemp=etemp*10+noe.number[i];
}
qf.pop();
qb.pop();
if(stemp==m)
{
cout<<nos.wal<<endl;
return ;
}
//r printf("%d %d\n",stemp,etemp);

data ne1,ne2;
ne1=nos,ne2=noe;
ne1.wal=nos.wal+1;
ne2.wal=noe.wal+1;
for(int i=0;i<4;i++)
{
for(int j=0;j<=9;j++)
{
ne1=nos;
ne2=noe;
if(i==0&&j==0) continue;
ne1.number[i]=j;
ne2.number[i]=j;
stemp=0,etemp=0;
for(int q=0;q<4;q++)
{
stemp=stemp*10+ne1.number[q];
etemp=etemp*10+ne2.number[q];
}

if(1000<=stemp&&etemp<=9999)
{
if(!vis[stemp]&&!is_prim[stemp])
{
vis[stemp]=1;
ne1.wal=nos.wal+1;
qf.push(ne1);
}
}
if(1000<=etemp&&etemp<=9999)
{
if(!vis[etemp]&&is_prim[etemp])
{
vis[etemp]=1;
ne2.wal=noe.wal+1;
qb.push(ne2);
}
}
}
}
}
}

int main(void)
{
prim_();
int t;
cin>>t;
while(t--)
{
init();
bfs();
}
}