POJ 3126 Prime Path（基础题）


Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11288		Accepted: 6398

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

#include <cstdio>
#include <map>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <queue>
#include <string>
using namespace std;
const int maxn=10000;
int vis[maxn],prime[maxn],mark[maxn];
int dis[maxn];
int st,ed;
int change(int a,int newd,int num)
{
int c[4];
int cnt=0;
while(a)
{
c[cnt++]=a%10;
a=a/10;
}
c[num]=newd;
int ans=(((c[3]*10+c[2])*10)+c[1])*10+c[0];
return ans;
}
void bfs()
{
memset(dis,0,sizeof(dis));
memset(mark,0,sizeof(mark));
queue<int>q;
q.push(st);
mark[st]=1;
while(!q.empty())
{
int temp=q.front();
q.pop();
for(int i=0;i<4;i++)
{
for(int j=0;j<=9;j++)
{
int newn=change(temp,j,i);
if(prime[newn]&&!mark[newn])
{
q.push(newn);
mark[newn]=1;
dis[newn]=dis[temp]+1;
if(newn==ed)
{
return;
}
}
}
}
}
}
int main()
{
int m=(int)sqrt(double(10000)+0.5);
memset(vis,0,sizeof(vis));
memset(prime,0,sizeof(prime));
for(int i=2;i<=m;i++)
{
if(!vis[i])
{
for(int j=i*i;j<=10000;j=j+i)
vis[j]=1;
}
}
for(int i=1000;i<=9999;i++)
{
if(!vis[i])
{
prime[i]=1;
}
}
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d %d",&st,&ed);
bfs();
printf("%d\n",dis[ed]);
}
return 0;
}