POJ 3126 Prime Path( 广搜 ) - c++编程基础

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POJ 3126 Prime Path( 广搜 )

2015-11-21 01:03:25 来源: 作者: 【大中小】浏览:2次

Prime Path

Time Limit: 1000MS	?	Memory Limit: 65536K
Total Submissions: 12974	?	Accepted: 7342

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

题目大意：给两个素数a，b（是4位数），问a是否能通过变换，变成b，变换原则：一次只能改变a的其中一位数字，并且转换后的数字必须也是素数，如1033能变成1733，但不能变成1233（因为1233不是素数），也不能变成3733（因为从1033到3733一次变换了2位数），若能，输出最少的变换次数，否则输出Impossible。
广搜，判断是否是素数可以先打表。

#include 
  
   
#include 
   
     #include 
    
      using namespace std; #define HUR 100 #define THO 1000 #define TEN 10 const int maxn=10000; bool prime[maxn+5]; int vis[maxn],s,e; void prime_table(){ int i,j; memset(prime,0,sizeof(prime)); for(i=2;i
     
       que; que.push(s); while(!que.empty()){ int t=que.front(); que.pop(); int d=t; d%=1000; for(i=1;i<10;i++){ int tt=d+i*THO; //变换千位 if(prime[tt]==0 && vis[tt]==0){ if(tt==e) return vis[t]; que.push(tt);vis[tt]=vis[t]+1; } } d=t%100+(t/1000*1000); for(i=0;i<10;i++){ int tt=d+i*HUR; //变换百位 if(prime[tt]==0 && vis[tt]==0){ if(tt==e) return vis[t]; que.push(tt);vis[tt]=vis[t]+1; } } d=t%10+t/100*100; for(i=0;i<10;i++){ int tt=d+i*TEN; //变换十位 if(prime[tt]==0 && vis[tt]==0){ if(tt==e) return vis[t]; que.push(tt);vis[tt]=vis[t]+1; } } d=t/10*10; for(i=0;i<10;i++){ int tt=d+i; //变换个位 if(prime[tt]==0 && vis[tt]==0){ if(tt==e) return vis[t]; que.push(tt);vis[tt]=vis[t]+1; } } } return 0; } int main() { int T,res; prime_table(); scanf("%d",&T); while(T--){ scanf("%d%d",&s,&e); if(s==e){ printf("0\n"); continue; } res=bfs(); if(res==0) printf("Impossible\n"); else printf("%d\n",res); } return 0; }


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