poj3264Balanced Lineup(RMQ)

http://poj.org/problem?id=3264

RMQ讲解 http://dongxicheng.org/structure/lca-rmq/

j = log2K

dp[i][j] = max(dp[i][j-1]+dp[i+(1<<(j-1))][j-1];

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<stdlib.h>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 #define N 50005
 int minz[N][],maxz[N][];
 void init(int n)
 {
     int i,j,o = floor(log10(double(n))/log10(double()));
     for(j =  ; j <= o ; j++)
         for(i =  ; i <= n+-(<<j) ; i++)
         {
             maxz[i][j] = max(maxz[i][j-],maxz[i+(<<(j-))][j-]);
             minz[i][j] = min(minz[i][j-],minz[i+(<<(j-))][j-]);
         }
 }
 int main()
 {
     int i,n,q,h;
     while(scanf("%d%d",&n,&q)!=EOF)
     {
         for(i = ; i <= n ; i++)
         {
             scanf("%d",&h);
             minz[i][] = h;
             maxz[i][] = h;
         }
         init(n);
         while(q--)
         {
             int a,b,t;
             scanf("%d%d",&a,&b);
             if(a>b)
             {
                 t = a;a = b; b = t;
             }
             int o = floor(log10(double(b-a+))/log10(double()));
             printf("%d\n",max(maxz[a][o],maxz[b-(<<o)+][o])-min(minz[a][o],minz[b-(<<o)+][o]));
         }
     }
     return ;
 }