POJ - 3126----Prime Path
2017-01-30 22:07
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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.
题目大意:
输入:给两个4位的素数,
操作:每次只能改变一位数字,且每次改变完后得到的数还是素数
输出:将前一个素数转化为后一个素数需要的步数。
思路:先将四位数的素数找出来存下,然后就简单的BFS即可
— 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
题目大意:
输入:给两个4位的素数,
操作:每次只能改变一位数字,且每次改变完后得到的数还是素数
输出:将前一个素数转化为后一个素数需要的步数。
思路:先将四位数的素数找出来存下,然后就简单的BFS即可
#include<iostream> #include<cstring> #include<queue> #include<cstdio> #include<cmath> using namespace std; const int MAX = 10000; struct vary{ int step;//记录步数 bool visit;//素数的访问标记 }; int prime[MAX]; vary A[MAX]; //素数制表 void isprime(void){ int i,j; for(i=1000 ; i<=MAX ; i++){ for(j=2 ; j<i ; j++){ if(i%j == 0){ break; } } if(j==i){ prime[i] = i; } } } int BFS(int start,int end){ queue<int> Q; memset(A,0,sizeof(A)); A[start].visit = true;//起始a已访问 Q.push(start); while(!Q.empty()){ int tem = Q.front(); int t[4];//分离数字 Q.pop(); //找到 if(tem == end){ cout<<A[tem].step<<endl; return 1; } t[0]=tem/1000; t[1]=tem%1000/100; t[2]=tem%100/10; t[3]=tem%10; //改变素数的一位数字入队 for(int i=0 ; i<4 ; i++){//要改的位 int k = t[i]; for(int j=0 ; j<10 ; j++){//要改的数 t[i] = j; int temp=t[0]*1000+t[1]*100+t[2]*10+t[3]; //与原数不同 未访问过 为素数 if(temp!=tem && !A[temp].visit && prime[temp]!=0){ A[temp].step = A[tem].step+1; A[temp].visit = true; Q.push(temp); } } t[i] = k;//还原该位; } } return 0; } int main(void){ int a,b,n; isprime();//预处理 cin>>n; for(int i=0 ; i<n ; i++){ cin>>a>>b; if(BFS(a,b) == 0){ cout<<"Impossible"<<endl; } } return 0; }
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