POJ3126Prime Path(AC)

Prime Path

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 14790 Accepted: 8341

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

31033 81791373 80171033 1033

Sample Output

670

#define _CRT_SECURE_NO_WARNINGS#include <stdio.h>int start = 0;int end = 0;int ans = 0;int num = 0; //一共有多少素数 （四位数的素数）int front = 0;int back = 0;int find = 0;/*第一次提交wa 看了discuss之后还是没有找到原因,后来百度发现原来还有要考虑找不到的输出Impossible的情况*//*第二次提交还是wa 发现竟然把素数的个数给打印出来了。。。Omg*//*第三次还是wa 发现素数表可能是求错了，个数不对，我的素数总是很多，这个不对的，原来自己写的是for (j = 2; (j*j) <= i; j++)还是写成for (j = 2; j <= (i/2); j++)然后ac的*/#define MAXINT 10001int map[MAXINT];typedef struct node{int prime;int step;}nodes;nodes que[MAXINT];int visit[MAXINT];//找出所有四位数的素数void getPrime(){int i = 0;int j = 0;int flag = 0;for (i = 1001; i < 9997; i++){flag = 0;for (j = 2; j <= (i/2); j++){if (0 == (i%j)){flag = 1;break;}}if (0 == flag){map[num++] = i;//printf("%d\n", i);}}//printf("num = %d\n", num);//一共有多少个素数return;}int getright(int index, int shu, int sp){int i = 0;int j = 0;int shuarry[4];int mude[4];int flag = 0;shuarry[3] = shu / 1000;shuarry[2] = (shu - shuarry[3] * 1000) / 100;shuarry[1] = (shu - shuarry[3] * 1000 - shuarry[2] * 100) / 10;shuarry[0] = shu - shuarry[3] * 1000 - shuarry[2] * 100 - shuarry[1] * 10;for (i = 0; i < num; i++){flag = 0;if (1 == visit[map[i]]) continue;mude[3] = map[i] / 1000;mude[2] = (map[i] - mude[3] * 1000) / 100;mude[1] = (map[i] - mude[3] * 1000 - mude[2] * 100) / 10;mude[0] = map[i] - mude[3] * 1000 - mude[2] * 100 - mude[1] * 10;if (shuarry[index] == mude[index]){continue;}for (j = 0; j < 4; j++){if (index == j) continue;if (shuarry[j] != mude[j]){flag = 1;break;}}//每一位是有很多的，不是单独一个的，这个很关键的if (0 == flag){que[back].prime = map[i];que[back++].step = sp + 1;visit[map[i]] = 1;}}return 0;}void bfs(){int i = 0;int search = 0;que[back].prime = start;que[back++].step = 0;visit[start] = 1;while (front < back){int x = que[front].prime;int sp = que[front].step;if (x == end){printf("%d\n", sp);find = 1; return;}//依次找和第i不同的素数,比如i=0,就找第一位不同，但是其他位都相同的素数for (i = 0; i < 4; i++){search = getright(i, x, sp);if (0 != search){}}front++;}return;}void init(){int i = 0;for (i = 0; i < MAXINT; i++){visit[i] = 0;que[i].prime = 0;que[i].step = 0;}front = 0;back  = 0;find  = 0;return;}int main(){int i = 0;int T = 0;freopen("input.txt", "r", stdin);scanf("%d", &T);init();for (i = 0; i < MAXINT; i++){map[i] = 0;}getPrime();for (i = 0; i<T; i++){ans = 0;init();scanf("%d %d", &start, &end);bfs();if (0 == find){printf("Impossible\n");}}return 0;}