# 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];

 1 #include <iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<stdlib.h>
5 #include<algorithm>
6 #include<cmath>
7 using namespace std;
8 #define N 50005
9 int minz[N][40],maxz[N][40];
10 void init(int n)
11 {
12     int i,j,o = floor(log10(double(n))/log10(double(2)));
13     for(j = 1 ; j <= o ; j++)
14         for(i = 1 ; i <= n+1-(1<<j) ; i++)
15         {
16             maxz[i][j] = max(maxz[i][j-1],maxz[i+(1<<(j-1))][j-1]);
17             minz[i][j] = min(minz[i][j-1],minz[i+(1<<(j-1))][j-1]);
18         }
19 }
20 int main()
21 {
22     int i,n,q,h;
23     while(scanf("%d%d",&n,&q)!=EOF)
24     {
25         for(i = 1; i <= n ; i++)
26         {
27             scanf("%d",&h);
28             minz[i][0] = h;
29             maxz[i][0] = h;
30         }
31         init(n);
32         while(q--)
33         {
34             int a,b,t;
35             scanf("%d%d",&a,&b);
36             if(a>b)
37             {
38                 t = a;a = b; b = t;
39             }
40             int o = floor(log10(double(b-a+1))/log10(double(2)));
41             printf("%d\n",max(maxz[a][o],maxz[b-(1<<o)+1][o])-min(minz[a][o],minz[b-(1<<o)+1][o]));
42         }
43     }
44     return 0;
45 }
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posted @ 2013-08-14 21:24  _雨  阅读(200)  评论(0编辑  收藏  举报