poj3126 Prime Path (广搜) 题解
Prime Path
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 15238 | Accepted: 8575 |
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.
— 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
Source
Northwestern Europe 2006
题目大意： 给定组数n，每组输入两个四位数的质数（也叫素数），问左边的质数最少经过可以几次变化可以变成后边质数？ 关键点来了：：怎么变呢？？ 是这样的：左边的质数每次只能更改个，十，百，千位上的一个，且每次更改完的新数字还必须是质数，比如上面说的 从1033---->8179 可以这样变：1033-->1733-->3733-->3739-->3779-->8779-->8179,这样就是经过了六次变动，且每次变动完的数字都是质数。。到这里你应该能想到每个位上的数字在一次变动中可能变大，也可能变小，，所以嘛～come on !上广搜！ 。。。。还有不要忘记判质数啊！！ 我的代码虽有点多，但灰常容易理解，不信请看
上代码：
#include <iostream>
#include<queue>
#include<string.h>
#include<cmath>
using namespace std;
int m,n,flag;
int step[11000],vis[10010],path[10010];
int check(int x)
{
int i;
for(i=2; i<=int(sqrt(double(x))); i++)
{
if(x%i==0)
return 0;
}
return 1;
}
void bfs(int x)
{
int s,i,j;
memset(step,0,sizeof(step));
memset(vis,0,sizeof(vis));
queue<int>q;
while(!q.empty())
q.pop();
q.push(x);
while(!q.empty())
{
x=q.front();
q.pop();
if(x==n)
{
flag=1;
return;
}
int a=x/1000;
int b=(x-a*1000)/100;
int c=(x-a*1000-b*100)/10;
int d=x-a*1000-b*100-c*10;
for(i=0; i<8; i++)
{
if(i==0)
{
for(j=1; j<10; j++)
{
if(a+j<10)
{
s=(a+j)*1000+b*100+c*10+d;
}
if(check(s)&&!vis[s])
{
vis[s]=1;
step[s]=step[x]+1;
path[s]=x;
q.push(s);
}
}
}
if(i==1)
{
for(j=1; j<10; j++)
{
if(b+j<10)
s=a*1000+(b+j)*100+c*10+d;
if(check(s)&&!vis[s])
{
vis[s]=1;
step[s]=step[x]+1;
path[s]=x;
q.push(s);
}
}
}
if(i==2)
{
for(j=1; j<10; j++)
{
if(b-j>=0)
s=a*1000+(b-j)*100+c*10+d;
if(check(s)&&!vis[s])
{
vis[s]=1;
step[s]=step[x]+1;
path[s]=x;
q.push(s);
}
}
}
if(i==3)
{
for(j=1; j<10; j++)
{
if(c+j<10)
s=a*1000+b*100+(c+j)*10+d;
if(check(s)&&!vis[s])
{
vis[s]=1;
step[s]=step[x]+1;
path[s]=x;
q.push(s);
}
}
}
if(i==4)
{
for(j=1; j<10; j++)
{
if(c-j>=0)
s=a*1000+b*100+(c-j)*10+d;
if(check(s)&&!vis[s])
{
vis[s]=1;
step[s]=step[x]+1;
path[s]=x;
q.push(s);
}
}
}
if(i==5)
{
for(j=1; j<10; j++)
{
if(d+j<10)
s=a*1000+b*100+c*10+d+j;
if(check(s)&&!vis[s])
{
vis[s]=1;
step[s]=step[x]+1;
path[s]=x;
q.push(s);
}
}
}
if(i==6)
{
for(j=1; j<10; j++)
{
if(d-j>=0)
s=a*1000+b*100+c*10+d-j;
if(check(s)&&!vis[s])
{
vis[s]=1;
step[s]=step[x]+1;
path[s]=x;
q.push(s);
}
}
}
if(i==7)
{
for(j=1; j<10; j++)
{
if(a-j>0)
s=(a-j)*1000+b*100+c*10+d;
if(check(s)&&!vis[s])
{
vis[s]=1;
step[s]=step[x]+1;
path[s]=x;
q.push(s);
}
}
}
}
}
}
int main()
{
int T,i,j;
cin>>T;
while(T--)
{
flag=0;
cin>>m>>n;
bfs(m);
if(flag==1)
{
cout<<step[n]<<endl;
}
else cout<<"Impossible"<<endl;
}
return 0;
}
题目大意： 给定组数n，每组输入两个四位数的质数（也叫素数），问左边的质数最少经过可以几次变化可以变成后边质数？ 关键点来了：：怎么变呢？？ 是这样的：左边的质数每次只能更改个，十，百，千位上的一个，且每次更改完的新数字还必须是质数，比如上面说的 从1033---->8179 可以这样变：1033-->1733-->3733-->3739-->3779-->8779-->8179,这样就是经过了六次变动，且每次变动完的数字都是质数。。到这里你应该能想到每个位上的数字在一次变动中可能变大，也可能变小，，所以嘛～come on !上广搜！ 。。。。还有不要忘记判质数啊！！ 我的代码虽有点多，但灰常容易理解，不信请看
上代码：
#include <iostream>
#include<queue>
#include<string.h>
#include<cmath>
using namespace std;
int m,n,flag;
int step[11000],vis[10010],path[10010];
int check(int x)
{
int i;
for(i=2; i<=int(sqrt(double(x))); i++)
{
if(x%i==0)
return 0;
}
return 1;
}
void bfs(int x)
{
int s,i,j;
memset(step,0,sizeof(step));
memset(vis,0,sizeof(vis));
queue<int>q;
while(!q.empty())
q.pop();
q.push(x);
while(!q.empty())
{
x=q.front();
q.pop();
if(x==n)
{
flag=1;
return;
}
int a=x/1000;
int b=(x-a*1000)/100;
int c=(x-a*1000-b*100)/10;
int d=x-a*1000-b*100-c*10;
for(i=0; i<8; i++)
{
if(i==0)
{
for(j=1; j<10; j++)
{
if(a+j<10)
{
s=(a+j)*1000+b*100+c*10+d;
}
if(check(s)&&!vis[s])
{
vis[s]=1;
step[s]=step[x]+1;
path[s]=x;
q.push(s);
}
}
}
if(i==1)
{
for(j=1; j<10; j++)
{
if(b+j<10)
s=a*1000+(b+j)*100+c*10+d;
if(check(s)&&!vis[s])
{
vis[s]=1;
step[s]=step[x]+1;
path[s]=x;
q.push(s);
}
}
}
if(i==2)
{
for(j=1; j<10; j++)
{
if(b-j>=0)
s=a*1000+(b-j)*100+c*10+d;
if(check(s)&&!vis[s])
{
vis[s]=1;
step[s]=step[x]+1;
path[s]=x;
q.push(s);
}
}
}
if(i==3)
{
for(j=1; j<10; j++)
{
if(c+j<10)
s=a*1000+b*100+(c+j)*10+d;
if(check(s)&&!vis[s])
{
vis[s]=1;
step[s]=step[x]+1;
path[s]=x;
q.push(s);
}
}
}
if(i==4)
{
for(j=1; j<10; j++)
{
if(c-j>=0)
s=a*1000+b*100+(c-j)*10+d;
if(check(s)&&!vis[s])
{
vis[s]=1;
step[s]=step[x]+1;
path[s]=x;
q.push(s);
}
}
}
if(i==5)
{
for(j=1; j<10; j++)
{
if(d+j<10)
s=a*1000+b*100+c*10+d+j;
if(check(s)&&!vis[s])
{
vis[s]=1;
step[s]=step[x]+1;
path[s]=x;
q.push(s);
}
}
}
if(i==6)
{
for(j=1; j<10; j++)
{
if(d-j>=0)
s=a*1000+b*100+c*10+d-j;
if(check(s)&&!vis[s])
{
vis[s]=1;
step[s]=step[x]+1;
path[s]=x;
q.push(s);
}
}
}
if(i==7)
{
for(j=1; j<10; j++)
{
if(a-j>0)
s=(a-j)*1000+b*100+c*10+d;
if(check(s)&&!vis[s])
{
vis[s]=1;
step[s]=step[x]+1;
path[s]=x;
q.push(s);
}
}
}
}
}
}
int main()
{
int T,i,j;
cin>>T;
while(T--)
{
flag=0;
cin>>m>>n;
bfs(m);
if(flag==1)
{
cout<<step[n]<<endl;
}
else cout<<"Impossible"<<endl;
}
return 0;
}