双向广搜 POJ 3126 Prime Path

Posted 2020-07-24 tech-chen

 

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\n");continue;
103         }
104         int temp=bfs();
105         if(temp==-1) printf("Impossible\n");
106         else printf("%d\n",temp);     
107      }
108     return 0;
109 }