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

时间:2015-09-14 22:44:44      阅读:403      评论:0      收藏:0      [点我收藏+]

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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\n", ans);
    }
}

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

 

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

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原文地址:http://www.cnblogs.com/andyqsmart/p/4808216.html

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