Prime Path
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 27852 | Accepted: 15204 |
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
大致题意：
给定两个四位素数a b，要求把a变换到b
变换的过程要保证 每次变换出来的数都是一个 四位素数，而且当前这步的变换所得的素数 与 前一步得到的素数 只能有一个位不同，而且每步得到的素数都不能重复。 不得有前导零！
求从a到b最少需要的变换次数。无法变换则输出Impossible
广度优先搜索
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
const int N = 10000;
bool visit[N];
bool Notprime[N];
int dist[N];
void GetNotPrimeVisit ( )
{
int i, j;
for( i=2; i<=N; i++ )
for( j=i<<1; j<=N; j+=i )
Notprime[j] = true;
fill( Notprime, Notprime+1000, true ); //以防万一把前一千个数去掉
}
int getnumber[45];
int x;
void SwapNumber( int v )
{
x = 0;
int i, a, b, c, d, n;
a = v / 1000;
b = v / 100 % 10;
c = v % 100 / 10;
d = v % 10;
n = v - a * 1000;
for( i=1; i<10; i++ )
getnumber[x++] = n + i * 1000;
n = v - b * 100;
for( i=0; i<10; i++ )
getnumber[x++] = n + i * 100;
n = v - c * 10;
for( i=0; i<10; i++ )
getnumber[x++] = n + i * 10;
n = v - d;
for( i=0; i<10; i++ )
getnumber[x++] = n + i;
}
queue <int> Q;
void BFS ( int a, int b )
{
int i, j, v;
int n = a;
Q.push( n );
visit[n] = true;
while( !Q.empty() )
{
v = Q.front();
if( v == b ) break;
Q.pop();
SwapNumber( v ); //得到40个入口
for( j=0; j<x; j++ )
{
i = getnumber[j];
if( !Notprime[i] && !visit[i] ) //使用素数表和visit剪枝
{
Q.push( i );
dist[i] = dist[v] + 1;
visit[i] = true;
}
}
}
if( !Q.empty() ) cout << dist[v] << endl;
else cout << "Impossible" << endl;
}
int main()
{
GetNotPrimeVisit(); //得到素数表
int t;
cin >> t;
while( t-- )
{
fill( visit, visit+N, false ); //初始化
fill( dist, dist+N, 0 ); //初始化
while( !Q.empty() ) Q.pop(); //初始化，清空队列
int a, b;
cin >> a >> b;
BFS( a, b ); //广搜
}
return 0;
}