poj3126--Prime Path(广搜)
Prime Path
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 11751 | Accepted: 6673 |
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
给出两个数s，e，都是素数。经过转化，将s转化为e的最小步数
规则，每一次仅仅能改动一位，每次得到的数都是素数。
素数筛跑出1000到10000内的全部素数。假设当中两个素数仅仅有一位不同。那么连接一条边。得到全部素数组合的图后用bfs直接搜索就能够
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
int a[11000] , check[11000] , tot ;
struct node{
int v , next ;
}p[2000000];
struct node1{
int u , t ;
};
queue <node1> que ;
int head[10000] , cnt , flag[10000] ;
void add(int u,int v)
{
p[cnt].v = v ;
p[cnt].next = head[u] ;
head[u] = cnt++ ;
}
void init()
{
memset(check,0,sizeof(check));
memset(head,-1,sizeof(head));
tot = cnt = 0 ;
int i , j , k , num ;
for(i = 2 ; i <= 10000 ; i++)
{
if( !check[i] )
a[tot++] = i ;
for(j = 0 ; j < tot ; j++)
{
if(i*a[j] >= 10000)
break;
check[i*a[j]] = 1 ;
if( i%a[j] == 0 )
break;
}
}
for(i = 0 ; i < tot ; i++)
if( (a[i]/1000) ) break;
k = i ;
for(i = k ; i < tot ; i++)
{
for(j = k ; j < i ; j++)
{
num = 0 ;
if( a[i]%10 != a[j]%10 )
num++ ;
if( a[i]/10%10 != a[j]/10%10 )
num++ ;
if( a[i]/100%10 != a[j]/100%10 )
num++ ;
if( a[i]/1000%10 != a[j]/1000%10 )
num++ ;
if(num == 1)
{
add(i,j);
add(j,i);
}
}
}
}
int find1(int x)
{
int low = 0 , mid , high = tot-1 ;
while(low <= high)
{
mid = (low+high)/2 ;
if(a[mid] == x)
return mid ;
else if(a[mid] < x)
low = mid + 1 ;
else
high = mid -1 ;
}
}
int bfs(int s,int e)
{
memset(flag,0,sizeof(flag));
while( !que.empty() )
que.pop();
int i , j , v ;
node1 low , high ;
low.u = s ;
low.t = 0 ;
flag[s] = 1 ;
que.push(low);
while( !que.empty() )
{
low = que.front();
que.pop();
if( low.u == e )
return low.t ;
for(i = head[low.u] ; i != -1 ; i = p[i].next)
{
v = p[i].v ;
if( !flag[v] )
{
flag[v] = 1;
high.u = v ;
high.t = low.t + 1 ;
que.push(high);
}
}
}
return 0;
}
int main()
{
int t , s , e ;
init();
scanf("%d", &t);
while(t--)
{
scanf("%d %d", &s, &e);
s = find1(s);
e = find1(e);
printf("%d\n", bfs(s,e) );
}
return 0;
}
