poj 3126 Prime Path（搜索专题）

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20237		Accepted: 11282

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

Source

Northwestern Europe 2006

方法很简单，但是我写代码还是写了好久，将思路转化成代码的速度还是太慢。

其中需要素数，又复习了一下素数筛法的方法。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

bool prime[10000];
int que[5000];
int step[5000];
int isprime[5000];
bool vis[10000];
int cou=0;

void getprime(){
    int num=10000;
    num=sqrt(num*1.0);
    memset(prime,true,sizeof(prime));
    prime[0]=prime[1]=false;
    for(int i=4;i<10000;i+=2){
        prime[i]=false;
    }
    for(int i=3;i<num;i+=2){
        if(prime[i]){
            for(int j=i*i;j<10000;j+=2*i){
                prime[j]=false;
            }
        }
    }
    for(int i=1000;i<10000;i++){
        if(prime[i]){
            isprime[cou++]=i;
        }
    }
}

bool judge(int a,int b){
    int sum=0;
    int t1[4],t2[4];
    for(int i=0;i<4;i++){
        t1[i]=a%10;
        a/=10;
        t2[i]=b%10;
        b/=10;
    }
    for(int i=0;i<4;i++){
        if(t1[i]==t2[i]){
            sum++;
        }
    }
    if(sum==3){
        return true;
    }else{
        return false;
    }
}

int bfs(int a,int b){
    int start=0;
    int endd=0;
    memset(vis,true,sizeof(vis));
    que[endd]=a;
    step[endd++]=0;
    while(start<endd){
        int now=que[start];
        int nowStep=step[start++];
        if(now==b){
            return nowStep;
        }
        for(int j=0;j<=cou;j++){
            int i=isprime[j];
            if(!vis[i]){
                continue;
            }
            if(prime[i]&&judge(i,now)){
                if(i==b){
                    return nowStep+1;
                }
                que[endd]=i;
                step[endd++]=nowStep+1;
                vis[i]=false;
            }
        }
    }
    return 0;
}

int main()
{
    int n;
    scanf("%d",&n);
    int a,b;
    getprime();
    for(int i=0;i<n;i++){
        scanf("%d %d",&a,&b);
        int ans=bfs(a,b);
        printf("%d\n",ans);
    }
    return 0;
}