POJ：3126-Prime Path

题目链接：http://poj.org/problem?id=3126

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Prime Path

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 25215 Accepted: 13889

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

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解题心得：

1. 题意就是一个长官的房间号是n（四位数），你每次可以改变n中的一个数，要你改变次数最少将n变成m，并且在改变工程中所有的数字都必须是一个素数，并且都是没有前导零的四位数，如果不能通过这样的改变达到m则输出-1;
2. 就是一个搜索加素数判断，直接bfs，只不过在入队的时候判断一下是否是一个素数就行了。

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#include <algorithm>
#include <stdio.h>
#include <queue>
#include <cstring>
using namespace std;
const int maxn = 1e4+100;
bool prim[maxn];
int n,m;

struct NODE {
    int va,step;
    NODE() {
        va = 0,step = 0;
    }
}now2,Next2;

void get_prim() {
    prim[0] = prim[1] = true;
    for(int i=2;i<maxn;i++) {
        if(prim[i])
            continue;
        for(int j=i*2;j<maxn;j+=i) {
            prim[j] = true;
        }
    }
}

int bfs() {
    bool vis[maxn];
    int now[6],Next[6];

    memset(vis,0, sizeof(vis));
    vis[n] = true;
    queue <NODE> qu;
    now2.va = n,now2.step = 0;
    qu.push(now2);

    while(!qu.empty()) {
        int temp = qu.front().va;
        int step = qu.front().step;
        qu.pop();

        if(temp == m)
            return step;

        int k = 0;
        while(k < 4) {
            now[k] = temp%10;
            temp /= 10;
            k++;
        }

        for(int i=0;i<4;i++) {
            for(int j=1;j<=9;j++) {
                for(int k=0;k<4;k++)
                    Next[k] = now[k];
                Next[i] = (now[i] + j) % 10;
                int num = 0;
                for(int k=3;k>=0;k--) {
                    num = num*10 + Next[k];
                }
                if(!prim[num] && !vis[num] && num >= 1000) {
                    Next2.step = step+1;
                    Next2.va = num;
                    qu.push(Next2);
                    vis[num] = true;
                }
            }
        }
    }
    return -1;
}

int main() {
    int t;
    scanf("%d",&t);
    get_prim();
    while(t--) {
        scanf("%d%d",&n,&m);
        int ans = bfs();
        if(ans != -1)
            printf("%d
",ans);
        else
            printf("Impossible
");
    }
    return 0;
}