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

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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 200005using 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;}