# bzoj 1636: [Usaco2007 Jan]Balanced Lineup

## 1636: [Usaco2007 Jan]Balanced Lineup

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 772  Solved: 560

## 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.

6 3

1

7

3

4

2

5

1 5

4 6

2 2

## Sample Input

* 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 Output

6

3

`#include<cstdio>#include <iostream>#define M 50010int max(int a,int b){return a>b?a:b;}int min(int a,int b){return a<b?a:b;}struct tree{int l,r,minn,maxx;}tr[4*M];int a[M];void make(int l,int r,int p){tr[p].l=l;tr[p].r=r;if(l==r){tr[p].minn=a[l];tr[p].maxx=a[l];return ;}int mid=(l+r)>>1;make(l,mid,p<<1);make(mid+1,r,p<<1|1);tr[p].minn=min(tr[p<<1].minn,tr[p<<1|1].minn);tr[p].maxx=max(tr[p<<1].maxx,tr[p<<1|1].maxx);}int fmin(int l,int r,int x){if(tr[x].l==l&&tr[x].r==r) return tr[x].minn;int mid=(tr[x].l+tr[x].r)>>1,q=x<<1;if(r<=mid) return fmin(l,r,q);else if(l>mid) return fmin(l,r,q+1);else return min(fmin(l,mid,q),fmin(mid+1,r,q+1));}int fmax(int l,int r,int x){if(tr[x].l==l&&tr[x].r==r) return tr[x].maxx;int mid=(tr[x].l+tr[x].r)>>1;if(r<=mid) return fmax(l,r,x<<1);else if(l>mid) return fmax(l,r,x<<1|1);else return max(fmax(l,mid,x<<1),fmax(mid+1,r,x<<1|1));}int main(){int n,m,i,x,y;scanf("%d%d",&n,&m);for(i=1;i<=n;i++) scanf("%d",&a[i]);make(1,n,1);for(i=0;i<m;i++){scanf("%d%d",&x,&y);printf("%d\n",fmax(x,y,1)-fmin(x,y,1));}}`

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