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poj 3126
Prime Path
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 11788 | Accepted: 6697 |
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.
1033The 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.
1733
3733
3739
3779
8779
8179
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
31033 81791373 80171033 1033
Sample Output
670
Source
Northwestern Europe 2006
#include <iostream>#include <cmath>#include <queue>using namespace std;const int MAXN = 10000;int prime1, prime2;struct node{ int num; int step;};bool used[MAXN];bool bPrime[MAXN];void IsPrime(){ bPrime[0] = bPrime[1] = 1; for(int i=2;i<=10000;i++) for(int j=i*2;j<=10000;j=j+i) bPrime[j]=1;}void bfs(){ queue <node> q; node p, t, head; while (!q.empty()) q.pop(); head.num = prime1; head.step = 0; used[prime1] = true; q.push(head); while (!q.empty()) { p = q.front(); q.pop(); if (p.num == prime2) { printf("%d\n", p.step); return; } int res, r, k; // 个位 res = p.num % 10; for (r = 0; r <= 9; ++r) { k = p.num - res + r; if (!used[k] &&! bPrime[k]) { t.num = k; t.step = p.step + 1; q.push(t); used[k] = true; } } // 十位 res = (p.num / 10) % 10; for (r = 0; r <= 9; ++r) { k = p.num - res * 10 + r * 10; if (!used[k] && !bPrime[k]) { t.num = k; t.step = p.step + 1; q.push(t); used[k] = true; } } // 百位 res = (p.num / 100) % 10; for (r = 0; r <= 9; ++r) { k = p.num - res * 100 + r * 100; if (!used[k] && !bPrime[k]) { t.num = k; t.step = p.step + 1; q.push(t); used[k] = true; } } // 千位 res = p.num / 1000; for (r = 0; r <= 9; ++r) { k = p.num - res * 1000 + r * 1000; if (p.num >= 1000) { if (!used[k] && !bPrime[k]) { t.num = k; t.step = p.step + 1; q.push(t); used[k] = true; } } } }}int main(){ int nCases; scanf("%d", &nCases); IsPrime(); while (nCases--) { scanf("%d%d", &prime1, &prime2); memset(used, false, sizeof(used)); bfs(); } return 0;}
poj 3126
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