[POJ]P3126 Prime Path[BFS]

[POJ]P3126

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 35230		Accepted: 18966

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

Source

Northwestern Europe 2006

---

两年没写了，现在已经真的菜爆了。

本题大致意思，给你两个四位的素数，一个是起始状态，另一个是终止状态，要使起始状态变为终止状态每一步可进行的操作为，将这个四位数的某一位更换，但要求新的数也必须是一个素数，问最少步数。

大致思路也比较简单，就是先欧拉线性筛把所有素数先筛出来，再用在线处理的方式bfs。（然而我bfs写炸了好几次...）

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int Maxn = 10010 ;
const int inf = 100000000 ;
namespace iNx{
    struct qq{
        int num;
        int c;
    };
    qq q[Maxn];
    int primer[Maxn],pos[Maxn];
    bool check[Maxn],exist[Maxn];
    int cnt,best;
    void Euler(){
        memset(check,1,sizeof check);
        int i,j;
        check[1]=false ;
        for(i=2;i<10000;i++){
            if(check[i]) primer[++cnt]=i;
            for(j=1;j<=cnt&&(i*primer[j]<10000);j++){
                check[i*primer[j]]=false ;
                if(i%primer[j]==0) break ;
            }
        }
    }
    int getn(int a,int i){
        if(i==1) return a%10;
        if(i==2) return (a/10)%10;
        if(i==3) return (a/100)%10;
        return a/1000;
    }
    void bfs(int a,int b){
        int head=1,tail=1;
        q[1].num=a;
        q[1].c=0;
        int i,j,h,k,m,t;
        while(tail>=head){
            h=q[head].num;
            if(h==b){
                printf("%d\n",q[head].c);
                break ;
            }
            exist[h]=true ;
            for(i=1;i<=4;i++){
                k=getn(h,i);
                for(j=1,m=1;j<i;j++) m*=10;
                for(j=0;j<=9;j++){
                    if(i==4&&j==0) continue ;
                    if(j==k) continue ;
                    t=h+j*m-k*m;
                    if(check[t]&&(!exist[t])){
                        q[++tail].num=t;
                        q[tail].c=q[head].c+1;
                    }
                }
            }
            head++;
        }
    }
    int main(){
        Euler();
        int n,a,b,i;
        scanf("%d",&n);
        for(i=1;i<=n;i++){
            scanf("%d%d",&a,&b);
            memset(exist,0,sizeof exist);
            bfs(a,b);
        }
        return 0;
    }
}
int main(){
    iNx::main();
    return 0;
}

现在要开始天天练习了，蒟蒻我太难了。