poj 3126 强行BFS

Prime Path
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 14758 Accepted: 8326
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

题意：
将一个四位素数经过几次变换变成另一个素数，如果不能输出0（变换方式为每次更改某一位数）
做法：
直接暴力即可，因为总共大约有10000个四位数，所以bfs解答树的节点最多为10000，所以不会超时。（PS：用了一下埃式筛法做素数表）

20min1A

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;

typedef struct{
    int v,h;
}Po;

int pr[10010];
int pp[10010];
queue<Po>qu;


void Ai(void){
    memset(pr,0,sizeof(pr));
    int i,j;
    for(i = 2;i<=10000;i++){
        if(!pr[i]){
            for(j = i*2;j<=10000;j += i) pr[j] = 1;
        }
    }
}

int bfs(int S,int E){
    while(!qu.empty()) qu.pop();
    Po a;
    a.v = S;
    a.h = 0;
    qu.push(a);
    memset(pp,0,sizeof(pp));
    pp[S] = 1;
    int i,j;
    while(!qu.empty()){
        Po term3 = qu.front();qu.pop();
        if(term3.v == E) return term3.h;
        int term = term3.v;
        int term1[4];
        term1[0] = (term/10)*10;
        term1[1] = (term/100)*100 + term%10;
        term1[2] = (term/1000)*1000 + term%100;
        for(i = 0;i<4;i++)
            switch(i){
            case 0:
                for(j = 0;j<10;j++){
                    int term2 = term1[0] + j;
                    if(!pp[term2]&&!pr[term2]){
                        pp[term2] = 1;
                        Po term4;
                        term4.v = term2;
                        term4.h = term3.h+1;
                        qu.push(term4);
                    }
                }
                break;
            case 1:
                for(j = 0;j<10;j++){
                    int term2 = term1[1] + j*10;
                    if(!pp[term2]&&!pr[term2]){
                        pp[term2] = 1;
                        Po term4;
                        term4.v = term2;
                        term4.h = term3.h+1;
                        qu.push(term4);
                    }
                }
                break;
            case 2:
                for(j = 0;j<10;j++){
                    int term2 = term1[2] + j*100;
                    if(!pp[term2]&&!pr[term2]){
                        pp[term2] = 1;
                        Po term4;
                        term4.v = term2;
                        term4.h = term3.h+1;
                        qu.push(term4);
                    }
                }
                break;
            case 3:
                for(j = 1;j<10;j++){
                    int term2 = term%1000 + j*1000;
                    if(!pp[term2]&&!pr[term2]){
                        pp[term2] = 1;
                        Po term4;
                        term4.v = term2;
                        term4.h = term3.h+1;
                        qu.push(term4);
                    }
                }
                break;
            }
    }
    return -1;
}

int main(){
    Ai();
    int n;
    scanf("%d",&n);
    while(n--){
        int S,E;
        scanf("%d%d",&S,&E);
        int k = bfs(S,E);
        if(-1 == k) printf("0\n");
        else printf("%d\n",k);
    }
    return 0;
}