POJ3264Balanced Lineup 线段树

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#线段树

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本文介绍了一种使用线段树的数据结构来优化查询特定范围内最大值和最小值的问题，该方法适用于处理大规模数据集中的区间查询任务，特别是对于高度等有上下限约束的游戏场景。

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 62002		Accepted: 28961
Case Time Limit: 2000MS

Description

For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.

Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.

Input

Line 1: Two space-separated integers, N and Q.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2..N+Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.

Output

Lines 1..Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.

Sample Input

Sample Output

6
3
0

AC代码：

//#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int MAXSIZE=50000+10;
const int inf=0x3f3f3f3f;
struct node{
    int value,value1;
    int rchild,lchild;
}tree[MAXSIZE*4];
int father[MAXSIZE];
int n,q;
void BuiltTree(int i,int l,int r){
    tree[i].lchild=l;
    tree[i].rchild=r;
    tree[i].value=-inf;
    tree[i].value1=inf;
    if(l==r){
        father[l]=i;
        return;
    }
    BuiltTree(i<<1,l,(l+r)/2);
    BuiltTree((i<<1)+1,(l+r)/2+1,r);
}
void Update(int i){
    if(i<=1) return;
    int fi=i>>1;
    int a=tree[fi<<1].value;
    int b=tree[(fi<<1)+1].value;
    int a1=tree[fi<<1].value1;
    int b1=tree[(fi<<1)+1].value1;
    tree[fi].value=max(a,b);
    tree[fi].value1=min(a1,b1);
    Update(i>>1);
}
int MAX,MIN;
void Query(int i,int l,int r){
    if(tree[i].lchild==l&&tree[i].rchild==r){
        MAX=max(MAX,tree[i].value);
        MIN=min(MIN,tree[i].value1);
        return;
    }
    i=i<<1;
    if(l<=tree[i].rchild){
        if(r<=tree[i].rchild) Query(i,l,r);
        else Query(i,l,tree[i].rchild);
    }
    i++;
    if(r>=tree[i].lchild){
        if(l>=tree[i].lchild) Query(i,l,r);
        else Query(i,tree[i].lchild,r);
    }
}
int main(){
    std::ios::sync_with_stdio(false);
    cin>>n>>q;
    BuiltTree(1,1,n);
    for(int i=1;i<=n;i++){
        cin>>tree[father[i]].value;
        tree[father[i]].value1=tree[father[i]].value;
        Update(father[i]);
    }
    int a,b;
    while(q--){
        cin>>a>>b;
        MAX=-inf;
        MIN=inf;
        Query(1,a,b);
        cout<<MAX-MIN<<endl;
    }
    return 0;
}