POJ3126Prime Path(AC)

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total 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

3
1033 8179
1373 8017
1033 1033

Sample Output

6
7
0

#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 10001

int 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;
}