Poj 3216 Prime Path

Prime Path

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 15126 Accepted: 8519

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

Source

Northwestern Europe 2006

题目大意：通过不停替换一位数字，并且替换完之后的数字为素数，求从给定数字到达另外给定数字的最小次数(最小数字代价，其实就是最小的次数，因为每次的数字都要重新购买).

思路：素数打表+BFS

AC 代码如下：注意只能用g++提交，用c++提交会RE，暂时不知道为什么，还望有知道的好心人告知。

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#include <queue>using namespace std;const int maxn=10000+5;int pp[maxn];void getPrim(){    memset(pp,0,sizeof(pp));    pp[0]=pp[1]=1;    for(int i=2;i<=sqrt(1.0*maxn);i++)        if(!pp[i])        for(int j=i*i;j<=maxn;j+=i)            pp[j]=1;}struct pt{    char x[5];    int cnt;};queue<pt> Q;void mysearch(int a,int b){    pt ta,tb;    int sign[maxn];    memset(sign,0,sizeof(sign));    while(!Q.empty()) Q.pop();    char x1[5], x2[5];    char tmp[5];    sprintf(x1,"%d",a);    sprintf(x2,"%d",b);    strcpy(ta.x,x1);    ta.cnt=0;    Q.push(ta);    while(!Q.empty()){        tb=Q.front();        //cout<<"tb"<<" "<<tb.x<<endl;        Q.pop();        if(strcmp(tb.x,x2)==0){            printf("%d\n",tb.cnt);            break;        }        for(int i=0;i<4;i++){            for(int j=0;j<=9;j++){                if(tb.x[i]==j+'0') continue;                if(j==0 && i==0) continue;                ta=tb;                ta.x[i]=j+'0';               // cout<<ta.x<<endl;                sscanf(ta.x,"%d",&a);                ta.cnt=tb.cnt+1;                //cout<<a<<endl;                if(!pp[a] &&!sign[a]){                    sign[a]=1;                    Q.push(ta);                }            }        }    }}int main(){    getPrim();    int n;    int a,b;    scanf("%d",&n);    while(n--){       scanf("%d%d",&a,&b);        if(a==b) printf("0\n");        else{            mysearch(a,b);        }    }    return 0;}