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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