POJ 3126 Prime Path （素数筛+bfs）

题目链接：POJ 3126

Describe:

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

题目大意：

给俩四位数的素数，每次只改变一位数字，问最少经过多少次变化能得到给定的素数。多个测试样例。

解题思路：

首先得到一张素数表，然后bfs，将与队首元素相差一位数并且是素数的四位数入队，直到得到最小值。

AC代码：

 

#include <iostream>
#include <queue>
#define N 9999
using namespace std;
int su[N],u[N],f[N],p[N]; // f用来标记是否走过，p用来记录变化次数
int prime()  // 玄学打表，这个cnt设置成全局变量竟然会变小，可能我的电脑中病毒了
{
    int cnt = 0; // 设置成局部变量
    for(int i = 2; i <= N; i++)
    {
        if(!u[i]) su[cnt++] = i;
        for(int j = 0; j < cnt; j++)
        {
            if(i*su[j] > N) break;
            u[i*su[j]] = 1;
            if(i%su[j] == 0) break;
        }
    }
    return cnt;
}
bool ok(int x, int y) // 判断两个四位数是否相差一位数
{
    int c = 0;
    while(x && y)
    {
        int t1 = x%10;
        int t2 = y%10;
        if(t1 != t2) c++;
        x /= 10;
        y /= 10;
    }
    if(c == 1) return true;
    else return false;
}
int main()
{
    int cnt = prime();
    int t,x,y,ans;
    cin >> t;
    while(t--)
    {
        queue<int> q;
        ans = 9999;
        for(int i = 0; i < N; i++) // 各种初始化
        {
            f[i] = 0;
            p[i] = 0;
        }
        cin >> x >> y;
        q.push(x); // 第一个元素入对
        f[x] = 1;
        while(!q.empty())
        {
            int tmp = q.front(); q.pop();
            if(tmp == y)
            {
                ans = min(ans,p[tmp]); // 得到最小值
                break;
            }
            for(int i = 0; i < cnt; i++)
            {
                if(su[i] > 1000 && f[su[i]] == 0 && ok(tmp,su[i])) // 过滤条件
                {
                    q.push(su[i]);
                    f[su[i]] = 1;
                    p[su[i]] = p[tmp]+1;
                }
            }
        }
        cout << ans << endl;
    }
    return 0;
}