POJ 3126 Prime Path（基础题）



Prime Path

Time Limit: 1000MS Memory Limit: 65536KTotal 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

31033 81791373 80171033 1033

Sample Output

670

#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;}