(线段树 区间查询)The Water Problem -- hdu -- 5443 （2015 ACM/ICPC Asia Regional Changchun Online）...

链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5443

 

The Water Problem

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 738    Accepted Submission(s): 591

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

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

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<vector>
#include<cmath>
using namespace std;

#define N 1110
#define INF 0xffffff
#define Lson r<<1
#define Rson r<<1|1
#define Mid  a[r].mid()

struct node
{
    int L, R, Max;
    int mid() {return (L+R)/2;}
}a[N*4];

int n, m;

void BuildTree(int r, int L, int R)
{
    a[r].L = L, a[r].R = R, a[r].Max = 0;

    if(L==R)
    {
        scanf("%d", &a[r].Max);
        return ;
    }

    BuildTree(Lson, L, Mid);
    BuildTree(Rson, Mid+1, R);

    a[r].Max = max(a[Lson].Max, a[Rson].Max);
}

int Query(int r, int L, int R)
{
    if(a[r].L==L && a[r].R==R)
    {
        return a[r].Max;
    }

    if(R<=Mid)
       return Query(Lson, L, R);
    else if(L>Mid)
       return Query(Rson, L, R);
    else
    {
        int LMax = Query(Lson, L, Mid);
        int RMax = Query(Rson, Mid+1, R);
        return max(LMax, RMax);
    }
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int i, L, R;

        scanf("%d", &n);

        BuildTree(1, 1, n);

        scanf("%d", &m);
        for(i=1; i<=m; i++)
        {
            scanf("%d%d", &L, &R);
            printf("%d\n", Query(1, L, R));
        }
    }
    return 0;
}

 

 

转载于:https://www.cnblogs.com/YY56/p/4820865.html