ACM学习历程—HDU 5443 The Water Problem(RMQ)（2015长春网赛1007题）

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,...,10^6} . 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

题目大意就是给定区间，求区间最值。

这里采用了RMQ的ST算法，直接套的模板。

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>

using namespace std;

const int maxN = 1005;
int n, q;
int a[maxN];
int ma[maxN][20];

void RMQ()
{
    memset(ma, 0, sizeof(ma));
    for (int i = 0; i < n; ++i)
        ma[i][0] = a[i];
    for (int j = 1; (1<<j) <= n; ++j)
        for (int i = 0; i+(1<<j)-1 < n; ++i)
            ma[i][j] = max(ma[i][j-1], ma[i+(1<<(j-1))][j-1]);
}

int query(int lt, int rt)
{
    int k = 0;
    while ((1<<(k+1)) <= rt-lt+1)
        k++;
    return max(ma[lt][k], ma[rt-(1<<k)+1][k]);
}

void input()
{
    scanf("%d", &n);
    for (int i = 0; i < n; ++i)
        scanf("%d", &a[i]);
    RMQ();
}

void work()
{
    scanf("%d", &q);
    int u, v, ans;
    for (int i = 0; i < q; ++i)
    {
        scanf("%d%d", &u, &v);
        ans = query(u-1, v-1);
        printf("%d
", ans);
    }
}

int main()
{
    //freopen("test.in", "r", stdin);
    int T;
    scanf("%d", &T);
    for (int times = 0; times < T; ++times)
    {
        input();
        work();
    }
    return 0;
}