poj 3126

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Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11788		Accepted: 6697

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

Sample Output

6
7
0

Source

Northwestern Europe 2006

#include <iostream>
#include <cmath>
#include <queue>
using namespace std;

const int MAXN = 10000;
int prime1, prime2;
struct node
{
        int num;
        int step;
};
bool used[MAXN];
bool bPrime[MAXN];

void IsPrime()
{
      bPrime[0] = bPrime[1] = 1;
      for(int i=2;i<=10000;i++)
          for(int j=i*2;j<=10000;j=j+i)
              bPrime[j]=1;
}

void bfs()
{
        queue <node> q;
        node p, t, head;
        while (!q.empty())
                q.pop();
        head.num = prime1;
        head.step = 0;
        used[prime1] = true;
        q.push(head);
        while (!q.empty())
        {
                p = q.front();
                q.pop();
                if (p.num == prime2)
                {
                        printf("%d\n", p.step);
                        return;
                }
                int res, r, k;
                // 个位
                res = p.num % 10;
                for (r = 0; r <= 9; ++r)
                {
                        k = p.num - res + r;
                        if (!used[k] &&! bPrime[k])
                        {
                                t.num = k;
                                t.step = p.step + 1;
                                q.push(t);
                                used[k] = true;
                        }
                }
                // 十位
                res = (p.num / 10) % 10;
                for (r = 0; r <= 9; ++r)
                {
                        k = p.num - res * 10 + r * 10;
                        if (!used[k] && !bPrime[k])
                        {
                                t.num = k;
                                t.step = p.step + 1;
                                q.push(t);
                                used[k] = true;
                        }
                }
                // 百位
                res = (p.num / 100) % 10;
                for (r = 0; r <= 9; ++r)
                {
                        k = p.num - res * 100 + r * 100;
                        if (!used[k] && !bPrime[k])
                        {
                                t.num = k;
                                t.step = p.step + 1;
                                q.push(t);
                                used[k] = true;
                        }
                }
                // 千位
                res = p.num / 1000;
                for (r = 0; r <= 9; ++r)
                {
                        k = p.num - res * 1000 + r * 1000;
                        if (p.num >= 1000)
                        {
                                if (!used[k] && !bPrime[k])
                                {
                                        t.num = k;
                                        t.step = p.step + 1;
                                        q.push(t);
                                        used[k] = true;
                                }
                        }
                }
        }
}

int main()
{
        int nCases;
        scanf("%d", &nCases);
        IsPrime();
        while (nCases--)
        {
                scanf("%d%d", &prime1, &prime2);
                memset(used, false, sizeof(used));
                bfs();
        }

        return 0;
}

poj 3126

标签：des blog http io ar os sp for strong

原文地址：http://www.cnblogs.com/a972290869/p/4101045.html

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