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
6 3
1
7
3
4
2
5
1 5
4 6
2 2
Sample Output
6
3
0
Source
題目大意：有N頭牛，對于第i頭牛，它的高度是h[i]；有Q個查詢，每次查詢給定一個區間[a,b]，求第a頭牛到第b頭牛中最高的和最矮的相差多少。對于這種典型的RMQ問題，可以建立一棵線段樹，然后查詢。
#include <iostream>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
const int MAXN = 50001;
struct segment
{
int left,right;
int high,low;
}tree[MAXN*2];
int t,s,height[MAXN];
void create(int l,int r,int index){
tree[index].left=l,tree[index].right=r;
if(l==r){
tree[index].high=height[l];
tree[index].low=height[l];
return;
}
int mid=(l+r)>>1;
create(l,mid,2*index);
create(mid+1,r,2*index+1);
tree[index].high=max(tree[2*index].high,tree[2*index+1].high);
tree[index].low=min(tree[2*index].low,tree[2*index+1].low);
}
void update(int l,int r,int index){
if(tree[index].left==l && tree[index].right==r){
if(tree[index].high>t) t=tree[index].high;
if(tree[index].low<s) s=tree[index].low;
return;
}
int mid=(tree[index].left+tree[index].right)>>1;
if(r<=mid)
update(l,r,2*index);
else if(l>mid)
update(l,r,2*index+1);
else{
update(l,mid,2*index);
update(mid+1,r,2*index+1);
}
}
int main(){
int i,n,q,x,y;
scanf("%d %d",&n,&q);
for(i=1;i<=n;i++) scanf("%d",&height[i]);
create(1,n,1);
for(i=1;i<=q;i++){
scanf("%d %d",&x,&y);
t=0,s=10000000;
update(x,y,1);
printf("%d\n",t-s);
}
return 0;
}