A - Balanced Lineup

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

题意就是找出区间内最大值最小值之差，刚学线段树，就拿这道题练了练手，代码有些繁琐。

看视频学的线段树，https://www.bilibili.com/video/av9350697?t=814

#include<stdio.h>
#include<iostream>
using namespace std;
struct node{
int left,right,value,small;//value最大  small最小
bool ex;//一开始做的时候最小值好多出现0  为了避免，当二叉树中存在该节点时才对其进行比较
};
int max(int x,int y)
{
    return x>y?x:y;
}
int min(int x,int y)
{

     if(x==0)
        return y;
    else if(y==0)
        return x;
    else
    return x<y?x:y;

}
node tree[200005];//50000  到二叉树里得开4*50000，完全二叉树2*50000
int  height [50005];
void build_tree(int i,int le,int ri)//建树 初始化 
{
    tree[i].left=le;
    tree[i].right=ri;
    tree[i].value=0;
    tree[i].small=0;
    tree[i].ex=true;
    if(le==ri)
    {
         tree[i].value=height[le];
        tree[i].small=height[le];

        return ;
    }

   build_tree(i<<1,le,(le+ri)>>1);
   build_tree((i<<1)+1,((le+ri)>>1)+1,ri);
}
int MAX;
int MIN;
void query(int i,int left,int right)//查找函数
{
   if(right==tree[i].right&&left==tree[i].left)
   {
       MAX=max(tree[i].value,MAX);
       MIN=min(tree[i].small,MIN);
       return ;
   }
   i=i<<1;
   if(left<=tree[i].right)
   {
       if(right<=tree[i].right)
        query(i,left,right);
       else
        query(i,left,tree[i].right);
   }
   i=i+1;
   if(right>=tree[i].left)
   {
    if(left>=tree[i].left)
      query(i,left,right);
    else
    query(i,tree[i].left,right);
   }
}
int main()
{
int N,Q;
int i;
int j,k;
cin>>N>>Q;

for(i=0;i<100005;i++)
    tree[i].ex=false;
for(i=1;i<=N;i++)
{
    scanf("%d",&height[i]);
}
build_tree(1,1,N);
for(i=4*N-1;i>=1;i=i-2)
{
    if(tree[i].ex==true&&tree[i-1].ex==true&&tree[i/2].ex==true&&(tree[i].small!=0||tree[i-1].small!=0))
    {
        tree[i/2].value=max( tree[i].value,tree[i-1].value);
       tree[i/2].small=min( tree[i].small,tree[i-1].small);

    }

}//因为只有二叉树叶子value和small有值，所以更新对他们父节点赋值
int x,y;


for(i=1;i<=Q;i++)
{
     MAX=0;
     MIN=10000005;
scanf("%d%d",&x,&y);
query(1,x,y);
printf("%d\n",-MIN+MAX);

}


    return 0;
}

POJ 3264 Balanced Lineup 线段树

A - Balanced Lineup

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