本文共 2302 字，大约阅读时间需要 7 分钟。

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

就是一个把求区间最大值与求最小值结合起来，我用了两次查询就过了。。。。

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #define INF 0x3f3f3f3f#define MAXN 200005#define Mod 10001using namespace std;#define lson l , m , rt << 1#define rson m + 1 , r , rt << 1 | 1const int maxn=222222;int MAX[maxn<<2],MIN[maxn<<2];void PushUp(int rt){ MAX[rt]=max(MAX[rt<<1],MAX[rt<<1|1]); MIN[rt]=min(MIN[rt<<1],MIN[rt<<1|1]);}void build(int l,int r,int rt){ if(r==l) { scanf("%d",&MAX[rt]); MIN[rt]=MAX[rt]; return; } int m=(l+r)>>1; build(lson); build(rson); PushUp(rt);}int queryma(int L,int R,int l,int r,int rt){ if(L<=l&&r<=R) { return MAX[rt]; } int m=(l+r)>>1; int ret=0; if(L<=m) ret=max(ret,queryma(L,R,lson)); if(R>m) ret=max(ret,queryma(L,R,rson)); return ret;}int querymi(int L,int R,int l,int r,int rt){ if(L<=l&&r<=R) { return MIN[rt]; } int m=(l+r)>>1; int ret=INF; if(L<=m) ret=min(ret,querymi(L,R,lson)); if(R>m) ret=min(ret,querymi(L,R,rson)); return ret;}int main(){ int n,q; while(~scanf("%d%d",&n,&q)) { build(1,n,1); while(q--) { int l,r; scanf("%d%d",&l,&r); printf("%d\n",queryma(l,r,1,n,1)-querymi(l,r,1,n,1)); } } return 0;}
           
          
         
        
       
      
     
    
   

转载地址：http://gpcvb.baihongyu.com/