hdu 5443(线段树水)

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 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

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

 

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
",MAXNUM);
        }
    }
}