HDU 5443 The Water Problem(水题找区间最大值)——2015 ACM/ICPC Asia Regional Changchun Online

本文讨论了在一个水资源稀缺的世界中，如何通过给定的水源大小序列和查询范围，找出最大水源的问题。文中提供了两种解决方案：一种是暴力搜索方法，通过遍历指定区间内的所有水源来找到最大的水源；另一种是利用线段树结构，预先存储每个区间的最大水源，从而在每次查询时以较低的时间复杂度找到答案。此外，文章还强调了数据结构在解决实际问题中的重要性。

The Water Problem

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

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,...,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

Source

2015 ACM/ICPC Asia Regional Changchun Online

/*********************************************************************/

题意：给你一个n个元素的序列，q次询问，每次询问让你找出区间[li,ri]中的最大值

解题思路：因为此题的数据规模比较小，一般想到的方法都是可以的，比如暴力呀、线段树呀，虽然一道水题没必要浪费时间写一个线段树，但在这里，我会把目前想到的这两种方法都讲一下

①首先是暴力的做法，这种做法想法比较简单，对于每次询问，直接遍历区间[li,ri]找出最大值就可以了

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 1005;
const int inf = 1000000000;
const int mod = 2009;
int s[N];
int main()
{
    int t,n,i,j,q,l,r,Max;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
            scanf("%d",&s[i]);
        scanf("%d",&q);
        for(i=0;i<q;i++)
        {
            scanf("%d%d",&l,&r);
            for(Max=0,j=l;j<=r;j++)
                Max=Max>s[j]?Max:s[j];
            printf("%d\n",Max);
        }
    }
    return 0;
}

②利用线段树，保留每个区间的最大值，这样对于每次询问，我们只需O(logn)的复杂度遍历线段树，找到最大值

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<deque>
#include<cmath>
#include<functional>
#include<iterator>
#include<set>
#include<utility>
#include<stack>
#include<queue>
#include<iostream>
using namespace std;
#define maxn 11000
#define mod 1000000007
int a[maxn];
struct node
{
    int l,r,ma;
}tree[maxn*4];
void build(int l,int r,int p)
{
    tree[p].l = l;
    tree[p].r = r;
    tree[p].ma = -1;
    if(l == r)
    {
        tree[p].ma = a[l];
        return;
    }
    int mid = (l+r)/2;
    build(l,mid,p*2);
    build(mid+1,r,p*2+1);
    tree[p].ma = max(tree[p*2].ma,tree[p*2+1].ma);
}
int query(int x,int y,int p)
{
    if(x == tree[p].l && y == tree[p].r)
        return tree[p].ma;
    int mid = (tree[p].l + tree[p].r)/2;
    if(x > mid)
        return query(x,y,p*2+1);
    else if(y <= mid)
        return query(x,y,p*2);
    else
        return max(query(x,mid,p*2),query(mid+1,y,p*2+1));

}
int main()
{
    int T,i,j,k,n,q,x,y;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;++i)
            scanf("%d",&a[i]);
        build(1,n,1);
        scanf("%d",&q);
        for(i=1;i<=q;++i)
        {
            scanf("%d%d",&x,&y);
            printf("%d\n",query(x,y,1));
        }

    }
    return 0;
}

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