hdu 1973 Prime Path 解题报告

Prime Path

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 301 Accepted Submission(s): 186

Problem 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
  
  
  
  

   
   解题代码
  
  
  
  

   
   
#include <iostream>
#include <stdio.h>
#include <cstring>
#include <queue>
using namespace std;
int m,n,flag;
int a[10001];
int vis[10001];
int d[4]= {1,10,100,1000};
struct node
{
    int x;
    int step;//记录当前转换的步数
};
void prime()//素数筛
{
    memset(a,0,sizeof(a));
    a[0]=a[1]=1;
    for(int i=2; i<=101; i++)
    {
        if(!a[i])
            for(int j=i*i; j<10001; j+=i)
                a[j]=1;
    }
}
queue<node> q;
int bfs(int s)
{
    while(!q.empty())//  多组测试用例，可能队列不为空，所以首先要清空队列
        q.pop();
    node st,dt;
    st.x=s;
    st.step=0;
    vis[st.x]=1;
    q.push(st);
    flag=0;
    while(!q.empty())
    {
        st=q.front();
        q.pop();
        if(st.x==n)
        {
            flag=1;
            return st.step;
        }
        for(int i=0; i<4; i++)
            for(int j=0; j<=9; j++)
            {
                if(i==3 && j==0)
                    continue;
                else
                {
                    if(j!=st.x/d[i]%10)//判断该位是不是和现有数字相同
                    {
                        int tx=((st.x/d[i]/10)*10+j)*d[i]+st.x%d[i];// 通项
                        if(!a[tx] && !vis[tx])
                        {
                            vis[tx]=1;
                            dt.x=tx;
                            dt.step=st.step+1;
                            q.push(dt);
                        }
                    }
                }
            }
    }
    return -1;
}
int main()
{
    int cs,count;
    scanf("%d",&cs);
    prime();
    while(cs--)
    {
        memset(vis,0,sizeof(vis));
        scanf("%d%d",&m,&n);
        if(m==n)
            printf("0\n");
        else
        {
            count=bfs(m);
            if(flag)
                printf("%d\n",count);
            else
                printf("Impossible\n");//注意找不着就输出Impossible
        }
    }
    return 0;
}