Balanced Lineup

最新推荐文章于 2022-01-04 21:21:11 发布

原创最新推荐文章于 2022-01-04 21:21:11 发布 · 242 阅读

0 ·

CC 4.0 BY-SA版权

线段树专栏收录该内容

12 篇文章

订阅专栏

本文介绍了一种使用线段树进行区间最大值与最小值查询的方法，适用于需要高效处理大量区间查询的问题场景。通过构建线段树，可以快速计算任意指定区间内元素的最大值和最小值，进而求得该区间内元素的高度差。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

对区间建树，维护区间最大值最小值。

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 43493		Accepted: 20419
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

Source

USACO 2007 January Silver

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#define LL long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int maxn=1<<30;
const int SIZE=5e4+10;
//int x[15]={0,31,30,31,30,31,30,31,31,30,31,30,31};
struct seg{
    int high,low;
}a[SIZE<<2];
void pushup(int rt){
    a[rt].high=max(a[rt<<1].high,a[rt<<1|1].high);
    a[rt].low=min(a[rt<<1].low,a[rt<<1|1].low);
}
void build(int l,int r,int rt){
    if(l==r){
        int x;
        scanf("%d",&x);
        a[rt].low=a[rt].high=x;
        return;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    pushup(rt);
}
void query(int &high,int &low,int L,int R,int l,int r,int rt){
    if(L<=l&&r<=R){
        if(a[rt].high>high)high=a[rt].high;
        if(a[rt].low<low)low=a[rt].low;
        return ;
    }
    int m=(l+r)>>1;
    if(L<=m)query(high,low,L,R,lson);
    if(R>m)query(high,low,L,R,rson);
}
int main()
{
    int n,q,x,y;
    while(scanf("%d%d",&n,&q)!=EOF){
        build(1,n,1);
        for(int i=0;i<q;i++){
            scanf("%d%d",&x,&y);
            int high=-maxn,low=maxn;
            query(high,low,x,y,1,n,1);
            printf("%d\n",high-low);
        }
    }
    return 0;
}