Poj 3264 balanced lineup

Description

For the daily milking, Farmer John'sNCows (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 shocould not differ too much in height.

Farmer John has made a listQ(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, NAnd 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 AAnd B(1 ≤ A≤ B≤ N), Representing the range of cows from ATo BIntrusive.

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

6 31734251 54 62 2

Sample output

630

A simple line segment tree question is that the line segment tree calculates the maximum and minimum values, and then the maximum and minimum values subtract from each other. You don't need to update them. You just need to update them when building them, simple.

#include<map>#include<set>#include<stack>#include<queue>#include<cmath>#include<vector>#include<cstdio>#include<string>#include<cstring>#include<cstdlib>#include<iostream>#include<algorithm>#define  inf 0x0f0f0f0fusing namespace std;const int maxn=50000+10;struct node{     int L,R,maxx,minx;}tree[maxn*4];int a[maxn];void build(int c,int x,int y){     tree[c].L=x; tree[c].R=y;     if (x==y)     {          tree[c].minx=a[x];          tree[c].maxx=a[x];          return;     }     int mid=x+(y-x)/2;     build(c*2,x,mid);     build(c*2+1,mid+1,y);     tree[c].maxx=max(tree[c*2].maxx,tree[c*2+1].maxx);     tree[c].minx=min(tree[c*2].minx,tree[c*2+1].minx);}int get_max(int c,int x,int y){     if (tree[c].L==x && tree[c].R==y)     return tree[c].maxx;     int mid=tree[c].L+(tree[c].R-tree[c].L)/2;     if (y<=mid) return get_max(c*2,x,y);     else if (x>mid) return get_max(c*2+1,x,y);     else return max(get_max(c*2,x,mid),get_max(c*2+1,mid+1,y));}int get_min(int c,int x,int y){     if (tree[c].L==x && tree[c].R==y)     return tree[c].minx;     int mid=tree[c].L+(tree[c].R-tree[c].L)/2;     if (y<=mid) return get_min(c*2,x,y);     else if (x>mid) return get_min(c*2+1,x,y);     else return min(get_min(c*2,x,mid),get_min(c*2+1,mid+1,y));}int get_diff(int x,int y){     return get_max(1,x,y)-get_min(1,x,y);}int main(){     int n,m,x,y;     while(scanf("%d%d",&n,&m)!=EOF)     {          for (int i=1;i<=n;i++) scanf("%d",&a[i]);          build(1,1,n);          while(m--)          {               scanf("%d%d",&x,&y);               printf("%d\n",get_diff(x,y));          }     }     return 0;}

Author: chensunrise