POJ-3126-Prime Path
2016-08-22 11:24
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Prime Path
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
给定两个四位素数a b,把a转换成b转换的过程要求每次转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。
输出从a到b最少需要的变换次数。无法变换则输出Impossible
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 16991 | Accepted: 9561 |
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
给定两个四位素数a b,把a转换成b转换的过程要求每次转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。
输出从a到b最少需要的变换次数。无法变换则输出Impossible
#include<iostream> #include<string.h> #include<stdio.h> #include<math.h> #include<algorithm> #include<queue> #include<vector> using namespace std; bool JudgePrim(int n) { if(n<2)return false; else { for(int i = 2; i<=sqrt(n); ++i) if(n%i==0)return false; return true; } } struct node { int data, s; } p, w; int f(int x, int i) { int k; if(i==0) k = x-(x%10); else if(i==1) k = x-x%100+x%10; else if(i==2) k =x-x%1000+x%100; else k = x%1000; return k; } int main() { int n, a, b, k; bool vis[10000]; scanf("%d", &n); while(n--) { scanf("%d%d", &a, &b); memset(vis, 0,sizeof(vis)); vis[a] = 1; queue<node>qq; p.data = a; p.s = 0; qq.push(p); while(!qq.empty()) { p = qq.front(); if(p.data==b) { printf("%d\n", p.s); break; } qq.pop(); for(int ii = 0; ii<4; ++ii) { for(int i = 0; i<=9; ++i) { if(i==0&&ii==3)continue; k = f(p.data, ii)+pow(10,ii)*i; if(!vis[k]&&JudgePrim(k)) { w.data = k; w.s = p.s+1; vis[w.data] = 1; qq.push(w); } } } } if(qq.empty())printf("Impossible\n"); } return 0; }
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