POJ 3126 math（BFS）

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 21581		Accepted: 11986

Descripti

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

Source

Northwestern Europe 2006

思路：对每位数字替换，直到成为目标数字。

代码：

#include "cstdio"
#include "iostream"
#include "algorithm"
#include "string"
#include "cstring"
#include "queue"
#include "cmath"
#include "vector"
#include "map"
#include "stdlib.h"
#include "set"
#define mj
#define db double
#define ll long long
using namespace std;
const int N=1e4+5;
const int mod=1e9+7;
const ll inf=1e16+10;
bool pri[N];
bool u[N],v[N];
int c[N];//统计步数
void init(){  
    int i,j;
    for(i=1000;i<=N;i++){
        for(j=2;j<i;j++)
            if(i%j==0){
                pri[i]=0;
                break;
            }
        if(j==i) pri[i]=1;
    }
}
int bfs(int s,int e){
    queue<int >q;
    memset(v,0, sizeof(v));
    memset(c,0, sizeof(c));
    int tmp,a[4],ans=s;
    q.push(s);
    v[s]=1;
    while(q.size()){
        int k=q.front();
        q.pop();
        a[0]=k/1000,a[1]=k/100%10,a[2]=k/10%10,a[3]=k%10;
        for(int i=0;i<4;i++){
            tmp=a[i];
            for(int j=0;j<10;j++){
                if(tmp!=j){
                    a[i]=j;
                    ans=a[0]*1000+a[1]*100+a[2]*10+a[3];
                    if(pri[ans]&&!v[ans]){//为素数且未使用过
                        c[ans]=c[k]+1;
                        v[ans]=1;
                        q.push(ans);
                    }
                    if(ans==e) {
                        printf("%d\n",c[ans]);
                        return 0;
                    }
                }
                a[i]=tmp;
            }
        }
        if(k==e) {
            printf("%d\n",c[k]);
            return 0;
        }
    }
    printf("Impossible\n");
    return 0;
}
int main()
{
    int n;
    int x,y;
    scanf("%d",&n);
    init();
    for(int i=0;i<n;i++){
        scanf("%d%d",&x,&y);
        bfs(x,y);
    }
    return 0;
}