POJ 3126 Prime Path( 廣搜 )
Prime Path
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 12974
Accepted: 7342
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
題目大意：給兩個素數a，b（是4位數），問a是否能通過變換，變成b，變換原則：一次只能改變a的其中一位數字，並且轉換後的數字必須也是素數，如1033能變成1733，但不能變成1233（因為1233不是素數），也不能變成3733（因為從1033到3733一次變換了2位數），若能，輸出最少的變換次數，否則輸出Impossible。
廣搜，判斷是否是素數可以先打表。
#include
#include
#include
using namespace std;
#define HUR 100
#define THO 1000
#define TEN 10
const int maxn=10000;
bool prime[maxn+5];
int vis[maxn],s,e;
void prime_table(){
int i,j;
memset(prime,0,sizeof(prime));
for(i=2;i que;
que.push(s);
while(!que.empty()){
int t=que.front();
que.pop();
int d=t;
d%=1000;
for(i=1;i<10;i++){
int tt=d+i*THO; //變換千位
if(prime[tt]==0 && vis[tt]==0){
if(tt==e) return vis[t];
que.push(tt);vis[tt]=vis[t]+1;
}
}
d=t%100+(t/1000*1000);
for(i=0;i<10;i++){
int tt=d+i*HUR; //變換百位
if(prime[tt]==0 && vis[tt]==0){
if(tt==e) return vis[t];
que.push(tt);vis[tt]=vis[t]+1;
}
}
d=t%10+t/100*100;
for(i=0;i<10;i++){
int tt=d+i*TEN; //變換十位
if(prime[tt]==0 && vis[tt]==0){
if(tt==e) return vis[t];
que.push(tt);vis[tt]=vis[t]+1;
}
}
d=t/10*10;
for(i=0;i<10;i++){
int tt=d+i; //變換個位
if(prime[tt]==0 && vis[tt]==0){
if(tt==e) return vis[t];
que.push(tt);vis[tt]=vis[t]+1;
}
}
}
return 0;
}
int main()
{
int T,res;
prime_table();
scanf("%d",&T);
while(T--){
scanf("%d%d",&s,&e);
if(s==e){
printf("0\n");
continue;
}
res=bfs();
if(res==0)
printf("Impossible\n");
else
printf("%d\n",res);
}
return 0;
}
