hdu 5443 ST表简单求最大最小值

最新推荐文章于 2022-11-20 18:24:37 发布

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最新推荐文章于 2022-11-20 18:24:37 发布

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分类专栏： rmq

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本文介绍了一种高效查询特定区间内最大水源大小的算法实现。针对一系列水源大小的数组，通过预处理建立数据结构，能够快速响应多次查询请求，找出指定范围内最大的水源值。此方法适用于水资源管理等场景。

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 $a_1, a_2, a_3, . . . , a_n$ 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 $a_l$ and $a_r$ .
Input
First you are given an integer $T (T \leq 10)$ indicating the number of test cases. For each test case, there is a number $n (0 \leq n \leq 1000)$ on a line representing the number of water sources. $n$ integers follow, respectively $a_1, a_2, a_3, . . . , a_n$ , and each integer is in $\{1, . . . , 10^6\}$ . On the next line, there is a number $q (0 \leq q \leq 1000)$ representing the number of queries. After that, there will be $q$ lines with two integers $l$ and $r (1 \leq l \leq r \leq n)$ 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

求 l r的最大值

#include <bits/stdc++.h>
using namespace std;
int a[1100];
int mx[1100][30];
int n;
void init()
{
    memset(mx,0,sizeof(mx));
    for(int i=1;i<=n;i++)
        mx[i][0]=a[i];
    int k=floor(log((double)n)/log(2.0));
    for(int i=1;i<=k;i++)
    {
        for(int j=n;j>=1;j--)
        {
            if(j+(1<<(i-1))<=n)
            {
                mx[j][i]=max(mx[j][i-1],mx[j+(1<<(i-1))][i-1]);
            }
        }
    }
}

int rmq(int i,int j)
{
    int k=floor(log((double)(j-i+1))/log(2.0));
    int maxx=max(mx[i][k],mx[j-(1<<k)+1][k]);
    return maxx;
}


int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        init();
        int m;
        scanf("%d",&m);
        while(m--)
        {
            int c,d;
            scanf("%d%d",&c,&d);
            printf("%d\n",rmq(c,d) );
        }
    }
}