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(简单) POJ 3126 Prime Path,BFS。

时间:2015-01-16 23:35:08      阅读:225      评论:0      收藏:0      [点我收藏+]

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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.
 
  题目就是说每一次变一个数,而且要求是素数,然后问最小变几次变成目标数,典型的BFS,先筛选出所有素数之后就直接爆搜就好。
 
代码如下:
技术分享
#include<iostream>
#include<cstring>
#include<queue>

using namespace std;

bool rem[10005];
int rans[10005];

void getPrime()
{
    for(int i=2;i<10000;++i)
        if(rem[i]==0)
            for(int j=i*2;j<10000;j+=i)
                rem[j]=1;    
}

int bfs(int S,int E)
{
    memset(rans,-1,sizeof(rans));

    queue<int> que;
    int t;
    int temp;

    que.push(S);
    rans[S]=0;

    while(!que.empty())
    {
        t=que.front();
        que.pop();

        if(t==E)
            return rans[E];

        for(int i=1;i<=9;++i)
        {
            temp=t%1000+i*1000;
            if(rans[temp]==-1&&rem[temp]==0)
            {
                rans[temp]=rans[t]+1;
                que.push(temp);
            }
        }
        for(int i=0;i<=9;++i)
        {
            temp=t%100+100*i+(t/1000)*1000;
            if(rans[temp]==-1&&rem[temp]==0)
            {
                rans[temp]=rans[t]+1;
                que.push(temp);
            }
        }
        for(int i=0;i<=9;++i)
        {
            temp=t%10+10*i+(t/100)*100;
            if(rans[temp]==-1&&rem[temp]==0)
            {
                rans[temp]=rans[t]+1;
                que.push(temp);
            }
        }
        for(int i=0;i<=9;++i)
        {
            temp=(t/10)*10+i;
            if(rans[temp]==-1&&rem[temp]==0)
            {
                rans[temp]=rans[t]+1;
                que.push(temp);
            }
        }
    }

    return -1;
}

int main()
{
    ios::sync_with_stdio(false);
    
    getPrime();

    int T;
    int a,b;
    int ans;

    cin>>T;

    while(T--)
    {
        cin>>a>>b;

        ans=bfs(a,b);

        if(ans==-1)
            cout<<"Impossible\n";
        else
            cout<<ans<<endl;
    }

    return 0;
}
View Code

 

(简单) POJ 3126 Prime Path,BFS。

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原文地址:http://www.cnblogs.com/whywhy/p/4229881.html

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