POJ - 3126 Prime Path
2018-03-13 20:49
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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
问从a到b需要经过几步变换,变换规则为,只能变成素数,并且只有一位数字变换
— 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 1033Sample Output
6 7 0
问从a到b需要经过几步变换,变换规则为,只能变成素数,并且只有一位数字变换
#include <iostream> #include <queue> #include<algorithm> #include<stdio.h> #include<string.h> using namespace std; #define MAXV 10000 bool prime[MAXV]; void init(){ int i,j; for(i=1000;i<=MAXV;i++){ for(j=2;j<i;j++) if(i%j==0){ prime[i]=false; break; } if(j==i) prime[i]=true; } } int bfs(int first,int last){ bool dis[MAXV]; queue <int>q; int v,i,j,temp,vtemp,count[MAXV],t[4]; memset(dis,false,sizeof(dis)); memset(count,0,sizeof(count)); q.push(first); dis[first]=true; while(!q.empty()){ v=q.front(); q.pop(); t[0]=v/1000; t[1]=v%1000/100; t[2]=v%100/10; t[3]=v%10; for(j=0;j<4;j++){ temp=t[j]; for(i=0;i<10;i++) if(i!=temp){ t[j]=i; vtemp=t[0]*1000+t[1]*100+t[2]*10+t[3]; if(!dis[vtemp] && prime[vtemp]){ count[vtemp]=count[v]+1; dis[vtemp]=true; q.push(vtemp); } if(vtemp==last) return count[vtemp]; } t[j]=temp; } if(v==last) return count[v]; } return -1; } int main(){ int n,a,b,key; init(); scanf("%d",&n); while(n--){ scanf("%d%d",&a,&b); key=bfs(a,b); if(key!=-1) printf("%d\n",key); else printf("Impossible\n"); } return 0; }
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