POJ 3126 Prime Path BFS

Posted 2022-11-03 cautx

题目：

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 35295		Accepted: 18998

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

题意:

给出两个素数下x,y，问:每次只改变一个数的情况使得x变成y所需要的最少步数。

分析:

给出的数只有四个位且首位不为0，以x为起点，从0->9枚举每一位数算出从x向外搜索的素数，更新从x到该素数的最短步数，枚举完从当前点出发的所有素数后从队列中取出新起点，直到队列为空或者找到了目标素数。

AC code:

#include<cstdio>
#include<cstring>
using namespace std;
struct node
    int k,step;
h[100005];
int s[10005];
bool u[10005];
int su[10005];
int num;
void olas()

    num=1;
    memset(u,true,sizeof(u));
    u[1]=false;
    for(int i=2;i<=10000;i++)
    
        if(u[i])    su[num++]=i;
        for(int j=1;j<num;j++)
        
            if(i*su[j]>10000)    break;
            u[i*su[j]]=false;
            if(i%su[j]==0)    break;
        
    

int change(int x,int i,int j)

    if(i==1)    return (x/10)*10+j;
    else if(i==2)    return (x/100)*100+j*10+x%10;
    else if(i==3)    return (x/1000)*1000+j*100+x%100;
    else if(i==4)    return j*1000+x%1000;

int main()

    //freopen("input.txt","r",stdin);
    olas();
    int t;
    scanf("%d",&t);
    while(t--)
    
        int ans=-1;
        int x,y;
        scanf("%d%d",&x,&y);
        h[1].k=x;
        h[1].step=0;
        int l=1,r=1;
        memset(s,100,sizeof(s));
        while(1)
        
            if(h[l].k==y)
            
                ans=h[l].step;
                break;
            
            int ts,tk;
            for(int i=1;i<=4;i++)
            
                for(int j=0;j<=9;j++)
                
                    if(!(j==0&&i==4))
                    
                        int tk=change(h[l].k,i,j);
                        if(!u[tk])    continue;
                        ts=h[l].step+1;
                        if(s[tk]<=ts)    continue;
                        if(tk==y)
                        
                            ans=ts;
                            break;
                        
                        s[tk]=ts;
                        r++;
                        h[r].k=tk;
                        h[r].step=ts;
                    
                
            
            if(l==r||ans>=0)    break;
            l++;
        
        if(ans>=0)    printf("%d\n",ans);
        else printf("Impossible\n");
    

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