Prime Path

Time Limit: 1000MS Memory Limit: 65536K

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

感覺題目略臭長。

給你兩個四位數的質數，問你從一個數變成另一個數需要多少步，不可能達到就輸出”Impossible”。

根據題目，每一步變化要遵循以下規則：
1.每次變化只能改變當前數中的其中一位數字；
2.每次變化得到的數都必須是質數。

於是又是一道思路十分耿直的bfs，無非是多了個素數篩打表。bfs每一步都查詢分別改變四位數字後可以得到的所有質數併入隊，每次出隊都判斷是否和所求數相等即可。

PS:前幾天在poj上寫這道題結果寫完還沒交poj就炸了……然後第二天發現國內oj炸了好多，都不知道去哪刷水題了，心塞……

#include <iostream>
#include <string>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <ctime>
#include <cmath>
#include <queue>
#include <map>
#define M 10005
#define INF 0x3f3f3f3f

using namespace std;

bool f[M], vis[M];

struct node
{
    int id, step;
}st, ed, tmp, t;

int g[4][3] = {{1000, 10000, 1000}, {100, 1000, 100}, {10, 100, 10}, {1, 10, 1}};

void prime()//素數篩打表
{
    for(int i = 2; i < M; i++)
        if(!f[i])
            for(int j = 2; i * j < M; j++)
                f[i * j] = true;
}

int bfs()
{
    memset(vis, true, sizeof(vis));
    queue<node> Q;
    Q.push(st);
    vis[st.id] = false;

    while(!Q.empty())
    {
        tmp = Q.front();
        if(tmp.id == ed.id) return tmp.step;
        Q.pop();
        t.step = tmp.step + 1;
        for(int i = 0; i < 4; i++)
        {
            for(int j = 0; j <= 9; j++)
            {
                if(i == 0 && j == 0)    continue;
                t.id = j * g[i][2] + tmp.id % g[i][0] + tmp.id / g[i][1] * g[i][1];
                if(!f[t.id] && vis[t.id])
                {
                    Q.push(t);
                    vis[t.id] = false;
                }
            }
        }
    }
    return -1;
}

int main()
{
    prime();
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d %d", &st.id, &ed.id);
        int ans = bfs();
        if(ans != -1)
            printf("%d\n", ans);
        else
            printf("Impossible\n");
    }
    return 0;
}

執行結果：