• 双向广搜 POJ 3126 Prime Path


     

    POJ 3126  Prime Path

    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 16204   Accepted: 9153

    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

    大致题意:

    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

     

    注意:双向广搜是在一个队列中实现的,只不过是交替进行罢了!

      1 #include<iostream>
      2 using namespace std;
      3 #include<cstdio>
      4 #include<cstring>
      5 #include<cmath>
      6 #include<cstdlib>
      7 #include<queue>
      8 #define N 10000
      9 struct prime{
     10     int c[4];
     11     int flag;
     12 };
     13 int dis[N];
     14 int visit[N];
     15 queue<prime>que;
     16 int js(int *m)
     17 {
     18     return (m[0]*1000+m[1]*100+m[2]*10+m[3]*1);
     19 }
     20 bool is_prime(int l)
     21 {
     22     bool flag=true;
     23     for(int i=2;i<=sqrt(l);++i)
     24     {
     25         if(l%i==0)
     26         {
     27             flag=false;
     28             break;
     29         }
     30     }
     31     return flag;
     32 }
     33 int bfs()
     34 {
     35     while(!que.empty())
     36     {
     37         prime x=que.front();
     38         que.pop();
     39         int now=js(x.c);
     40         for(int i=1;i<=9;++i)
     41         {
     42                 prime nx=x;
     43                 nx.c[0]=i;
     44                 int shu=js(nx.c);
     45                 if(!visit[shu]&&is_prime(shu))
     46                 {
     47                     visit[shu]=x.flag;
     48                     nx.flag=x.flag;
     49                     que.push(nx);
     50                     dis[shu]=dis[now]+1;
     51                 }
     52                 else if(visit[shu]&&visit[shu]!=x.flag)
     53                 {
     54                     return dis[now]+dis[shu]+1;
     55                 }
     56         }
     57         for(int j=1;j<=3;++j)
     58         {
     59             for(int i=0;i<=9;++i)
     60           {  
     61                 prime nx=x;
     62                 nx.c[j]=i;
     63                 int shu=js(nx.c);
     64                 if(!visit[shu]&&is_prime(shu))
     65                 {
     66                     visit[shu]=x.flag;
     67                     nx.flag=x.flag;
     68                     que.push(nx);
     69                     dis[shu]=dis[now]+1;
     70                 }
     71                 else if(visit[shu]&&visit[shu]!=x.flag)
     72                 {
     73                     return dis[now]+dis[shu]+1;
     74                 }
     75             }
     76             
     77         }
     78     }
     79     return -1;
     80 }
     81 int main()
     82 {
     83     int tex;
     84     scanf("%d",&tex);
     85     char a[10];
     86     while(tex--)
     87     {
     88         while(!que.empty()) que.pop();
     89         memset(dis,0,sizeof(dis));
     90         memset(visit,0,sizeof(visit));
     91         scanf("%s",a);
     92         que.push(prime{a[0]-'0',a[1]-'0',a[2]-'0',a[3]-'0',1});
     93         int p=(a[0]-'0')*1000+(a[1]-'0')*100+(a[2]-'0')*10+(a[3]-'0');
     94         visit[p]=1;dis[p]=0;
     95         int q=p;
     96         scanf("%s",a);
     97         que.push(prime{a[0]-'0',a[1]-'0',a[2]-'0',a[3]-'0',2});
     98         p=(a[0]-'0')*1000+(a[1]-'0')*100+(a[2]-'0')*10+(a[3]-'0');
     99         visit[p]=2;dis[p]=0;
    100         if(q==p) 
    101         {
    102             printf("0
    ");continue;
    103         }
    104         int temp=bfs();
    105         if(temp==-1) printf("Impossible
    ");
    106         else printf("%d
    ",temp);     
    107      }
    108     return 0;
    109 }
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  • 原文地址:https://www.cnblogs.com/c1299401227/p/5575841.html
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