POJ - 3126 - Prime

先上题目

Prime Path
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 9259   Accepted: 5274

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

　 题意比较简单，就是给你两个4位的素数，问你能不能经过有限的步骤将第一个素数转化为第二个素数，其中转化的要符合一下的要求：每一次只可以改变这个四位数的某一位，首位不能为0，每一次转化以后得到的数要还是素数。如果不可以转化得到目标素数，输出Impossible。
　 做法也是比较简单，先筛出10000以内的素数出来，然后枚举当前的这个素数可以转化的其他素数，然后进行bfs，当遇到目标素数的时候就输出步数；如果转化不了就会把可以转化的素数都转化了一遍，最后循环就会结束。这里为了能够判断是否能转化，需要标记它转化过哪些素数，转化过的素数如果再出现，就不需要再搜索这个数了。



上代码：

  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <queue>
  4 #include <map>
  5 #define LL long long
  6 #define MAX (10000+10)
  7 using namespace std;
  8 
  9 bool f[MAX],M[MAX];
 10 
 11 typedef struct
 12 {
 13     int l;
 14     int st;
 15 }S;
 16 queue<S> q;
 17 
 18 void dedeal()
 19 {
 20     LL i,j,n;
 21     n=MAX;
 22     for(i=2;i<=n;i++)
 23     {
 24         if(!f[i])
 25         {
 26             for(j=i*i;j<=n;j+=i)  f[j]=1;
 27         }
 28     }
 29 }
 30 
 31 int check(int x,int y)
 32 {
 33     int t,i;
 34     S d;
 35     d.l=x;
 36     d.st=0;
 37     memset(M,0,sizeof(M));
 38     while(!q.empty()) q.pop();
 39     q.push(d);
 40     M[d.l]=1;
 41     while(!q.empty())
 42     {
 43         d=q.front();
 44         q.pop();
 45         if(d.l==y) return d.st;
 46         d.st++;
 47         x=d.l;
 48         t=x/10*10;
 49         for(i=0;i<10;i++)
 50         {
 51             d.l=t+i;
 52             if(!f[d.l] && d.l!=x && !M[d.l])
 53             {
 54                 M[d.l]=1;
 55                 q.push(d);
 56             }
 57         }
 58         t=x/100*100+x%10;
 59         for(i=0;i<100;i+=10)
 60         {
 61             d.l=t+i;
 62             if(!f[d.l] && d.l!=x && !M[d.l])
 63             {
 64                 M[d.l]=1;
 65                 q.push(d);
 66             }
 67         }
 68         t=x/1000*1000+x%100;
 69         for(i=0;i<1000;i+=100)
 70         {
 71             d.l=t+i;
 72             if(!f[d.l] && d.l!=x && !M[d.l])
 73             {
 74                 M[d.l]=1;
 75                 q.push(d);
 76             }
 77         }
 78         t=x%1000;
 79         for(i=1000;i<10000;i+=1000)
 80         {
 81             d.l=t+i;
 82             if(!f[d.l] && d.l!=x && !M[d.l])
 83             {
 84                 M[d.l]=1;
 85                 q.push(d);
 86             }
 87         }
 88     }
 89     return -1;
 90 }
 91 
 92 int main()
 93 {
 94     int n,c,x,y;
 95     //freopen("data.txt","r",stdin);
 96     dedeal();
 97     scanf("%d",&n);
 98     while(n--)
 99     {
100         scanf("%d %d",&x,&y);
101         c=check(x,y);
102         if(c==-1) printf("Impossible\n");
103         else printf("%d\n",c);
104     }
105     return 0;
106 }
3126

 

posted @ 2013-08-01 20:08  海拉鲁的林克  阅读(218)  评论(0)  编辑  收藏  举报