意甲冠軍 ?給你兩個(gè)4位質(zhì)數(shù)a, b ?每次你可以改變a個(gè)位數(shù)，但仍然需要素?cái)?shù)的變化 ?乞討a有多少次的能力，至少修改成b

基礎(chǔ)的bfs ?注意數(shù)的處理即可了 ?出隊(duì)一個(gè)數(shù) ?然后入隊(duì)全部能夠由這個(gè)素?cái)?shù)經(jīng)過(guò)一次改變而來(lái)的素?cái)?shù) ?知道得到b

#include <cstdio> #include <cstring> using namespace std; const int N = 10000; int p[N], v[N], d[N], q[N], a, b;void initPrime() {memset(v, 0 , sizeof(v));for(int i = 2; i * i < N; ++i)if(!v[i]) for(int j = i; i * j < N; ++j) v[i * j] = 1;for(int i = 2; i < N ; ++i) p[i] = !v[i]; }int bfs() {int c, t, le = 0, ri = 0;memset(v, 0, sizeof(v));q[ri++] = a, v[a] = 1, d[a] = 0;while(le < ri){c = q[le++];if( c == b) return d[c];for(int i = 1; i < N; i *= 10){for(int j = 0; j < 10; ++j) //把c第i數(shù)量級(jí)的數(shù)改為j{if(i == 1000 && j == 0) continue;t = c / (i * 10) * i * 10 + i * j + c % i;if(p[t] && !v[t])v[t] = 1, d[t] = d[c] + 1, q[ri++] = t;}}}return -1; }int main() {int cas;scanf("%d", &cas);initPrime();while(cas--){scanf("%d%d", &a, &b);if((a = bfs()) != -1) printf("%d\n", a);else puts("Impossible");}return 0; } 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

3 1033 8179 1373 8017 1033 1033

Sample Output

6 7 0

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轉(zhuǎn)載于:https://www.cnblogs.com/zfyouxi/p/4841927.html

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