hdu 5443 The Water Problem(RMQ区间最值)

题目：http://acm.hdu.edu.cn/showproblem.php?pid=5443

Problem Description

In Land waterless, water is a very limited resource. People always fight for the biggest source of water. Given a sequence of water sources with a1,a2,a3,...,an representing the size of the water source. Given a set of queries each containing 2 integers l and r , please find out the biggest water source between al and ar .

 

Input

First you are given an integer T(T≤10) indicating the number of test cases. For each test case, there is a number n(0≤n≤1000) on a line representing the number of water sources. n integers follow, respectively a1,a2,a3,...,an , and each integer is in {1,...,106} . On the next line, there is a number q(0≤q≤1000) representing the number of queries. After that, there will be q lines with two integers l and r(1≤l≤r≤n) indicating the range of which you should find out the biggest water source.

 

Output

For each query, output an integer representing the size of the biggest water source.

 

Sample Input

   
   
   
   

    
    
    
    3
1
100
1
1 1
5
1 2 3 4 5
5
1 2
1 3
2 4
3 4
3 5
3
1 999999 1
4
1 1
1 2
2 3
3 3
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    100
2
3
4
4
5
1
999999
999999
1
   
   
   
   

分析：最开始不知道RMQ算法，用线段树做的，代码写的挺长。现在了解了RMQ区间最值算法，重新写一遍。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn=1e3+10;
int maxsum[maxn][20]; //minsum[maxn],
void RMQ(int sum){
    for(int j=1;j<20;j++){
        for(int i=1;i<=sum;i++){
            if(i+(1<<j)-1<=sum){
                maxsum[i][j]=max(maxsum[i][j-1],maxsum[i+(1<<(j-1))][j-1]);
                //minsum[i][j]=min(maxsum[i][j-1],minsum[i+(1<<(j-1))][j-1]);
            }
        }
    }
}
int main()
{
    //freopen("cin.txt","r",stdin);
    int t,sum,a,q;
    cin>>t;
    while(t--){
        scanf("%d",&sum);
        for(int i=1;i<=sum;i++){
            scanf("%d",&a);
            //minsum[i][0]=
            maxsum[i][0]=a;
        }
        RMQ(sum);
        scanf("%d",&q);
        int l,r;
        while(q--){
            scanf("%d%d",&l,&r);
            int k=(int)log2(r-l+1.0);
            printf("%d\n",max(maxsum[l][k],maxsum[r-(1<<k)+1][k]));
        }
    }
    return 0;
}