COJ 1090: Number Transformation (bfs)

Description

 In this problem, you are given a pair of integers A and B. You can transform any integer number A to B by adding x to A.This x is an integer number which is a prime below A.Now,your task is to find the minimum number of transformation required to transform S to another integer number T.

Input

 Input contains multiple test cases.Each test case contains a pair of integers S and T(0< S < T <= 1000) , one pair of integers per line. 

Output

 For each pair of input integers S and T you should output the minimum number of transformation needed as Sample output in one line. If it's impossible ,then print 'No path!' without the quotes.

Sample Input

5 7

3 4

Sample Output

Need 1 step(s)

No path!

先用筛法得到1000以内的所有素数，然后bfs即可。

View Code

#include <stdio.h>

#include <string.h>

#include <queue>

using namespace std;

#define N 1001

int prime[N],cnt;

int step[N];

bool b[N];

int s,t;

void filter()

{

    int i,j;

    for(i=2;i<N;i++)

    {

        if(b[i])

        {

            prime[cnt++]=i;

            for(j=i<<1;j<N;j+=i)    b[j]=false;

        }

    }

}

void init()

{

    memset(b,true,sizeof(b));

    cnt=0;

}

void bfs()

{

    queue<int> q;

    bool success=false;

    int ans,cur,next;

    memset(step,-1,sizeof(step));

    q.push(s);

    step[s]=0;

    while(!q.empty() && !success)

    {

        cur=q.front();

        q.pop();

        if(cur==t)

        {

            success=true;

            ans=step[cur];

        }

        for(int i=0;prime[i]<cur && !success;i++)

        {

            next=cur+prime[i];

            if(next>t)  continue;

            if(step[next]==-1)

            {

                step[next]=step[cur]+1;

                if(next==t)

                {

                    success=true;

                    ans=step[next];

                }

                else    q.push(next);

            }

        }

    }

    if(success) printf("Need %d step(s)\n",ans);

    else    puts("No path!");

}

int main()

{

    init();

    filter();

    while(~scanf("%d%d",&s,&t)) bfs();

    return 0;

}