hdoj 5443 The Water Problem 【RMQ】

最新推荐文章于 2020-08-23 13:49:15 发布

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本文介绍了一种解决区间最大值查询（RMQ）问题的有效算法。通过预处理数组，可以在O(log n)时间内回答任意区间内的最大值。适用于水问题等应用场景。

The Water Problem

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 343 Accepted Submission(s): 281

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,...,anrepresenting 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

Sample Output

RMQ裸题

ac代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define MAXN 10010
#define INF 1000000000
using namespace std;
int A[MAXN];
int N, M;
int Amax[MAXN][50];
void RMQ_init()
{
    for(int i = 1; i <= N; i++)
        Amax[i][0] = A[i];
    for(int j = 1; (1<<j) <= N; j++)
    {
        for(int i = 1; i + (1<<j)-1 <= N; i++)
            Amax[i][j] = max(Amax[i][j-1], Amax[i+(1<<(j-1))][j-1]);
    }
}
int query(int L, int R)
{
    int k = 0;
    while((1<<(k+1)) <= R-L+1) k++;
    return max(Amax[L][k], Amax[R-(1<<k)+1][k]);
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &N);
        for(int i = 1; i <= N; i++)
            scanf("%d", &A[i]);
        RMQ_init();
        scanf("%d", &M);
        int x, y;
        while(M--)
        {
            scanf("%d%d", &x, &y);
            printf("%d\n", query(x, y));
        }
    }
    return 0;
}