题目背景
题目描述：
每天,农夫 John 的N(1 <= N <= 50,000)头牛总是按同一序列排队. 有一天, John 决定让一些牛们玩一场飞盘比赛. 他准备找一群在对列中为置连续的牛来进行比赛. 但是为了避免水平悬殊,牛的身高不应该相差太大. John 准备了Q (1 <= Q <= 200,000) 个可能的牛的选择和所有牛的身高 (1 <= 身高 <= 1,000,000). 他想知道每一组里面最高和最低的牛的身高差别.
输入：
第1行：N,Q
第2到N+1行：每头牛的身高
第N+2到N+Q+1行：两个整数A和B，表示从A到B的所有牛。（1<=A<=B<=N）
输出：
输出每行一个数，为最大数与最小数的差
题目描述
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.
一个农夫有N头牛，每头牛的高度不同，我们需要找出最高的牛和最低的牛的高度差。
输入格式
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.
输出格式
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.
输入输出样例
6 3 1 7 3 4 2 5 1 5 4 6 2 2
6 3 0
看n<=50000，必定是用O（nlogn）时间复杂度的RMQ。
但RMQ的实现方式有许多种，树状数组、线段树、ST表都可以。
鉴于前面的大佬只给出了线段树与ST表做法，我就来一发常数小、代码短的树状数组，据说速度一交就排rank3，180ms，还是很快的。
用树状数组原理同时维护区间最大值最小值即可。
#include <cstdio> #include <iostream> #include <cstring> #include <algorithm> using namespace std; int n,m; int op,l,r,x; int a[2000005]; inline int read(){ int s=0,w=1; char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();} while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar(); return s*w; } struct note{ int left; int right; int sign; int flag1; int flag2; }tree[2000005]; int pushup(int last) { tree[last].flag1=max(tree[last*2].flag1,tree[last*2+1].flag1); tree[last].flag2=min(tree[last*2].flag2,tree[last*2+1].flag2); return 0; } int pushdown(int last){ if(tree[last*2].right!=tree[last*2].left) tree[last*2].sign+=tree[last].sign; if(tree[last*2+1].right!=tree[last*2+1].left) tree[last*2+1].sign+=tree[last].sign; tree[last*2].flag1+=tree[last].sign; tree[last*2].flag2+=tree[last].sign; tree[last*2+1].flag1+=tree[last].sign; tree[last*2+1].flag2+=tree[last].sign; tree[last].sign=0; } int buildtree(int last,int left,int right) { tree[last].left=left; tree[last].right=right; if(left==right) return tree[last].flag1=tree[last].flag2=a[left]; int mid=(left+right)/2; buildtree(last*2,left,mid); buildtree(last*2+1,mid+1,right); pushup(last); } int repair(int last) { int left=tree[last].left,right=tree[last].right; if(l<=left&&right<=r) { tree[last].flag1+=x; tree[last].flag2+=x; tree[last].sign+=x; return 0; } if(tree[last].sign) pushdown(last); int mid=(left+right)/2; if(l<=mid) repair(last*2); if(mid<r) repair(last*2+1); pushup(last); return 0; } int check1(int last) { int left=tree[last].left,right=tree[last].right; if(l<=left&&right<=r) { return tree[last].flag1; } if(tree[last].sign) pushdown(last); int mid=(left+right)/2; int Max1=0,Max2=0; if(l<=mid) Max1=check1(last*2); if(mid<r) Max2=check1(last*2+1); return max(Max1,Max2); } int check2(int last){ int left=tree[last].left,right=tree[last].right; if(l<=left&&right<=r){ return tree[last].flag2; } if(tree[last].sign) pushdown(last); int mid=(left+right)/2; int Min1=0X3f3f3f3f,Min2=0X3f3f3f3f; if(l<=mid) Min1=check2(last*2); if(mid<r) Min2=check2(last*2+1); return min(Min1,Min2); } int main(){ n=read(); m=read(); for(int i=1;i<=n;i++){ a[i]=read(); } buildtree(1,1,n); for(int i=1;i<=m;i++){ l=read(); r=read(); printf("%d\n",(check1(1)-check2(1))); } return 0; }