The Water Problem RMQ区间最大值

The Water Problem

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2249    Accepted Submission(s): 1757

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 witha1,a2,a3,...,an representing the size of the water source. Given a set of queries each containing2 integers l and r, please find out the biggest water source between al and ar.

 

Input

First you are given an integerT(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

3110011 151 2 3 4 551 21 32 43 43 531 999999 141 11 22 33 3

 

Sample Output

1002344519999999999991

 

Source

2015 ACM/ICPC Asia Regional Changchun Online

 

Recommend

hujie

区间最大值  RMQ算法预处理即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n;
int dp[50010][50];
void rmq()
{
    for(int j=1; (1<<j)<=n; j++)
    {
        for(int i=1; i+(1<<j)-1<=n; i++)
        {
            dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
        }
    }
}
int rmq1(int l,int r)
{


    int k=0;
    while((1<<(k+1))<=r-l+1) k++;
    return max(dp[l][k],dp[r-(1<<k)+1][k]);
}


int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int j=1; j<=n; j++)
        {
            scanf("%d",&dp[j][0]);
        }
        rmq();
        int q;
        scanf("%d",&q);
        while(q--)
        {


            int a,b;
            scanf("%d%d",&a,&b);
            printf("%d\n",rmq1(a,b));
        }
    }
    return 0;
}

人一我百！人十我万！永不放弃~~~怀着自信的心，去追逐梦想。