意甲冠军 给你两个4位质数a, b 每次你可以改变a个位数,但仍然需要素数的变化 乞讨a有多少次的能力,至少修改成b
基础的bfs 注意数的处理即可了 出队一个数 然后入队全部能够由这个素数经过一次改变而来的素数 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数量级的数改为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 ", a); else puts("Impossible"); } return 0; }
Prime Path
Description
![](http://poj.org/images/3126_1.jpg)
— 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
3
1033 8179
1373 8017
1033 1033
Sample Output
6
7
0
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