• HDU


    Prime Path

    Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 987    Accepted Submission(s): 635


    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
     
    Source
     
    Recommend
    wangye
     
     
    题意:给你两个四位数,都是素数,每次改变可以改变其中的任何一个数字,但要求改变后的四位数(没有前导零)依然是素数,问最少改变几次可以使得第一个数改为第二个数
    思路:可以先用埃氏筛法把1000-9999的所有的素数都选出来,之后就四个位数,每个位数最多改变八次,就搜索一下,我开了isprime和step两个数组,当然也可以开一个结构体,但只改变一个数字需要花点功夫,我是枚举了个位,十位,百位,千位。
     
    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cstring>
    #include<cmath>
    #include<cstdlib>
    #include<queue>
    #include<set>
    #include<vector>
    using namespace std;
    #define INF 0x3f3f3f3f
    #define eps 1e-10
    #define PI acos(-1.0)
    #define ll long long
    int const maxn = 9999;
    const int mod = 1e9 + 7;
    int gcd(int a, int b) {
        if (b == 0) return a;  return gcd(b, a % b);
    }
    
    bool isprime[maxn];
    int step[maxn];
    bool vis[maxn];
    int num1,num2;
    void getprime(int n)
    {
        for(int i=0;i<=n;i++)
            isprime[i]=true;
        isprime[0]=isprime[1]=false;
        for(int i=2;i<=n;i++)
        {
            for(int j=2*i;j<=n;j=j+i)
                isprime[j]=false;
        }
    }
    
    int bfs(int st )
    {
     
        vis[st]=true;
        step[st]=0;
        queue<int>que;
        que.push(st);
        while(que.size())
        {
            int p=que.front();
            que.pop();
            if(p == num2)
            {
                return step[num2];
                break;
            }
            int ge=p % 10;
            int shi=(p/10)%10;
            int bai=(p/100)%10;
            int qian=p/1000;
            for(int i=0;i<=9;i++)
            {
                int next;
                if(i!=ge)
                {
                    next=p-ge+i;
                    if(next>=1000 && next<=9999 && isprime[next] && vis[next]==false)
                    {
                        step[next]=step[p]+1;
                        que.push(next);
                        vis[next]=true;
                        //   cout<<"ge";
                    }
                }
                if(i!=shi)
                {
                    next=p-shi*10+i*10;
                    if(next>=1000 && next<=9999 && isprime[next] && vis[next]==false)
                    {
                        step[next]=step[p]+1;
                        que.push(next);
                        vis[next]=true;
                    //    cout<<"shi";
                    }
                }
                if(i!=bai)
                {
                    next=p-bai*100+i*100;
                    if(next>=1000 && next<=9999 && isprime[next] && vis[next]==false)
                    {
                        step[next]=step[p]+1;
                        que.push(next);
                        vis[next]=true;
                    //    cout<<"bai";
                    }
                }
                if(i!=qian)
                {
                    next=p-qian*1000+i*1000;
                    if(next>=1000 && next<=9999 && isprime[next] && vis[next]==false)
                    {
                        step[next]=step[p]+1;
                        que.push(next);
                        vis[next]=true;
                   //     cout<<"qian";
                    }
                }
            }
                
        }
        return -1;
    }
    int main()
    {
        int t;
        scanf("%d",&t);
        getprime(9999);
        while(t--)
        {
            memset(vis,false,sizeof(vis));
            memset(step,0,sizeof(step));
            scanf("%d %d",&num1,&num2);
            
    //        for(int i=2;i<=9999;i++)
    //            if(isprime[i])
    //                cout<<i<<" ";
            int ans=bfs(num1);
            if(ans>=0)
              printf("%d
    ",ans);
            else
                printf("Impossible
    ");
        }
    }
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  • 原文地址:https://www.cnblogs.com/smallhester/p/9567606.html
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