POJ 3126 Prime Path( 广搜 )
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.
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
题目大意：给两个素数a，b（是4位数），问a是否能通过变换，变成b，变换原则：一次只能改变a的其中一位数字，并且转换后的数字必须也是素数，如1033能变成1733，但不能变成1233（因为1233不是素数），也不能变成3733（因为从1033到3733一次变换了2位数），若能，输出最少的变换次数，否则输出Impossible。
广搜，判断是否是素数可以先打表。
#include <stdio.h>
#include <queue>
#include <string.h>
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<maxn;i++)
if(!prime[i])
for(j=i*i;j<maxn;j+=i)
prime[j]=1;
}
int bfs(){
int i;
memset(vis,0,sizeof(vis));
vis[s]=1;
queue<int> 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; //变换千位
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; //变换百位
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; //变换十位
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; //变换个位
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;
}