CSUOJ 1299 - Number Transformation II 打表预处理水DP

http://122.207.68.93/OnlineJudge/problem.php?id=1299

第二个样例解释..

3 6

3->4->6..两步..

由此可以BFS也可以DP..但关键是要离线把100000内每个数的约数情况预先处理出来..否则会超时...

Program:

 

#include<iostream>
#include<stdio.h>
#include<cmath>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#define pi acos(-1)
#define ll long long 
#define oo 2139062143
#define MAXN 200005
using namespace std;
struct node
{
       int next;
}h[MAXN];
int n,m,dp[MAXN],p[3000000][2];
int main()
{
       int x,i,k;
       m=0;
       memset(h,0,sizeof(h));
       memset(p,0,sizeof(p));
       for (i=1;i<=MAXN;i++)
          for (x=i*2;x<=MAXN;x+=i)
          {
                p[++m][0]=i;
                p[m][1]=h[x].next;
                h[x].next=m;        
          } 
       while (~scanf("%d%d",&n,&m))
       {  
             memset(dp,0x7f,sizeof(dp));
             dp[n]=0;
             for (x=n;x<=m;x++) 
             {
                   k=h[x].next;
                   while (k)
                   {
                        i=p[k][0];
                        if (x+i<=m)
                           if (dp[x+i]==-1 || dp[x+i]>dp[x]+1)
                              dp[x+i]=dp[x]+1;
                        k=p[k][1];
                   }
             }
             if (dp[m]==oo) printf("sorry, can not transform\n");
               else       printf("%d\n",dp[m]);
       }
       return 0;
}