首页 > 代码库 > poj3126

poj3126

2024-10-18 02:08:02   210人阅读

题目链接：

http://poj.org/problem?

id=3126

题目：

Prime Path
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 10737   Accepted: 6110

Description

技术分享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
这道题目思想非常easy 。。可是实现起来比較复杂  。。

。

用bfs搜索。。

。

还有就是假设预处理打素数表的话会快许多。。

。

。可是我打的素数表很很的戳。。

。

。。可是还是0MS.。。

代码例如以下:

#include<cstdio>
#include<queue>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn=9999+10;
int prime[maxn];
bool vis[maxn];
struct point
{
    int x;
    int time;
};
point start,end;
bool is_prime(int x)
{
   for(int i=2;i<=sqrt(x*1.0);i++)
   {
       if(x%i==0)
        return false;
   }
   return true;
}
void init()
{
    for(int i=1000;i<=maxn;i++)
        prime[i]=i;
    for(int i=1000;i<=maxn;i++)
    {
        if(is_prime(prime[i]))
            prime[i]=1;
        else
            prime[i]=0;
    }
}
int bfs()
{
     queue<point>q;
     point old,current;
     q.push(start);
     while(!q.empty())
     {
        old=q.front();
        q.pop();
        if(old.x==end.x)
            return old.time;
        for(int i=1;i<=9;i=i+2)//个位
        {
            current.x=old.x/10*10+i;
            if(current.x==old.x)
                continue;
          //  printf("%d\n",current.time);
            if(prime[current.x]&&!vis[current.x])
            {
                vis[current.x]=1;
                current.time=old.time+1;
                q.push(current);
            }
        }
        for(int i=0;i<=9;i++)
        {
             current.x=old.x/100*100+i*10+old.x%10;
             if(current.x==old.x)
                continue;
           // printf("%d\n",current.time);
            if(prime[current.x]&&!vis[current.x])
                {
                    vis[current.x]=1;
                    current.time=old.time+1;
                    q.push(current);
                }
        }
        for(int i=0;i<=9;i++)
        {
            int tmp1=old.x%100;
            current.x=(old.x/1000*10+i)*100+tmp1;
            if(current.x==old.x)
                continue;
            if(prime[current.x]&&!vis[current.x])
                {
                    vis[current.x]=1;
                    current.time=old.time+1;
                    q.push(current);

                }
        }
        for(int i=1;i<=9;i++)
        {
            int tmp1=old.x%1000;
            current.x=i*1000+tmp1;
            if(current.x==old.x)
                continue;

            if(prime[current.x]&&!vis[current.x])
            {
                vis[current.x]=1;
                current.time=old.time+1;//printf("%d\n",current.time);
                q.push(current);
            }
        }
    }
    return -1;
}
int main()
{
    memset(prime,0,sizeof(prime));

    int T;
    scanf("%d",&T);
    init();
    while(T--)
    {
        memset(vis,false,sizeof(vis));
        scanf("%d %d",&start.x,&end.x);
        start.time=0;
        int ans=bfs();
        //printf("nas:%d\n",ans);
        if(ans!=-1)
             printf("%d\n",ans);
         else
             printf("Impossible\n");
    }
    return 0;
}


poj3126


联系
我们