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最新推荐文章于 2025-09-07 20:32:33 发布

转载最新推荐文章于 2025-09-07 20:32:33 发布 · 66 阅读

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原文链接：http://www.cnblogs.com/liyinggang/p/5568312.html

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#java

本文介绍了一种解决区间最大值查询问题的有效方法。通过构建一种特殊的数据结构，可以在多个查询中快速找到指定范围内的最大值，适用于水资源分配等场景。

The Water Problem

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1308 Accepted Submission(s): 1038

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

Input

First you are given an integer

Output

For each query, output an integer representing the size of the biggest water source.

Sample Input

Sample Output

区域赛水题，区间最值。。

//单点更新+区间查找
#include<iostream>
#include <stdio.h>
#include <string.h>
using namespace std;

const int Max = 1500;
int MAXNUM;
int a[Max];
struct Tree{
    int Max;
    int r,l;
}t[4*Max];
int MAX(int k,int j){
    if(k>=j) return k;
    return j;
}
void build(int idx,int l,int r){
    t[idx].l = l;
    t[idx].r=r;
    if(l==r){
        t[idx].Max = a[l];
        return;
    }
    int mid = (l+r)>>1;
    build(idx<<1,l,mid);
    build(idx<<1|1,mid+1,r);
    t[idx].Max = MAX(t[idx<<1].Max,t[idx<<1|1].Max); //父亲节点

}
void query(int idx,int l,int r,int L,int R){
    if(l>=L&&r<=R) {
        MAXNUM = MAX(MAXNUM,t[idx].Max);
        return;
    }
    int mid = (l+r)>>1;
    if(mid>=L)
    query(idx<<1,l,mid,L,R);
    if(mid<R)
    query(idx<<1|1,mid+1,r,L,R);
}

int main()
{
    int tcase;
    scanf("%d",&tcase);
    while(tcase--){
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        build(1,1,n);
        int m;
        scanf("%d",&m);
        for(int i=1;i<=m;i++){
            int l,r;
            scanf("%d%d",&l,&r);
            MAXNUM = -999999999;
            query(1,1,n,l,r);
            printf("%d\n",MAXNUM);
        }
    }
}