[HDU 1973]--Prime Path(BFS,素数表)

时间：2023-01-30 08:35:25

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1973

Prime Path

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem 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
 



Source
 
 
题目大意：给定两个四位素数a  b，要求把a变换到b变换的过程要保证  每次变换出来的数都是一个 四位素数，而且当前这步的变换所得的素数
　　　　　与前一步得到的素数  只能有一个位不同，而且每步得到的素数都不能重复。求从a到b最少需要的变换次数。无法变换则输出Impossible
 
解题思路：打一个素数表，然后基于每个数的每一位bfs搜索即可，具体的可见代码~~~
 
代码如下：
 #include <iostream>

 #include <queue>

 #include <cstdio>

 #include <cstring>

 using namespace std;

 struct node{

     int cur, step;

 }now, Next;

 int vis[], star, finish, prime[] = { , ,  };

 void init(){

     for (int i = ; i < ; i++){

         if (!prime[i]){

             for (int j = ; i*j < ; j++)

                 prime[i*j] = ;

         }

     }

 }

 int bfs(){

     queue<node> Q;

     vis[star] = ;

     now.cur = star, now.step = ;

     Q.push(now);

     while (!Q.empty()){

         int i, j;

         char num[];

         now = Q.front();

         Q.pop();

         if (now.cur == finish) return now.step;

         for (i = ; i < ; i++){

             sprintf(num, "%d", now.cur);

             for (j = ; j < ; j++){

                 if (j ==  && i == )

                     continue;

                 if (i == )

                     Next.cur = j *  + (num[] - '') *  + (num[] - '') *  + (num[] - '');

                 else if (i == )

                     Next.cur = j *  + (num[] - '') *  + (num[] - '') *  + (num[] - '');

                 else if (i == )

                     Next.cur = j *  + (num[] - '') *  + (num[] - '') *  + (num[] - '');

                 else if (i == )

                     Next.cur = j + (num[] - '') *  + (num[] - '') *  + (num[] - '') * ;

                 if (!prime[Next.cur] && !vis[Next.cur])

                 {

                     Next.step = now.step + ;

                     vis[Next.cur] = ;

                     Q.push(Next);

                 }

             }

         }

     }

     return -;

 }

 int main(){

     int t, ans;

     cin >> t;

     init();

     while (t--){

         cin >> star >> finish;

         memset(vis, , sizeof(vis));

         ans = bfs();

         if (ans == -) cout << "Impossible\n";

         else cout << ans << endl;

     }

     return ;

 }

其实这道题学校OJ(Swust OJ)也有但是坑爹的后台数据变成a+b的后台数据了，Orz~~~(无爱了)

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