poj3264 最大值与最小值的差

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：就是取最大值与最小值的差。
题解：就是线段树的基本功能里的。

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cmath>
 5 #include<cstring>
 6 const int MAXN=50007;
 7 char s[20];
 8 int a[MAXN]={0},tree[MAXN*4]={0},fuck[MAXN*4]={0},t,num,x,n,y,q;
 9 using namespace std;
10 void build(int l,int r,int p)
11 {
12     if (l==r) tree[p]=a[l];
13     else
14     {
15         int mid=(l+r)>>1;
16         build(l,mid,p*2);
17         build(mid+1,r,p*2+1);
18         tree[p]=max(tree[p*2],tree[p*2+1]);
19     } 
20 }
21 void build_fuck(int l,int r,int p)
22 {
23     if (l==r) fuck[p]=a[l];
24     else
25     {
26         int mid=(l+r)>>1;
27         build_fuck(l,mid,p*2);
28         build_fuck(mid+1,r,p*2+1);
29         fuck[p]=min(fuck[p*2],fuck[p*2+1]);
30     } 
31 }
32 int query(int l,int r,int p,int x,int y)
33 {
34     if (l==x&&r==y) return tree[p];
35     else
36     {
37         int mid=(l+r)>>1;
38         if (y<=mid) return query(l,mid,p*2,x,y);
39         else if(x>=mid+1) return query(mid+1,r,p*2+1,x,y);
40             else return max(query(l,mid,p*2,x,mid),query(mid+1,r,p*2+1,mid+1,y));
41     }
42 }
43 int query_fuck(int l,int r,int p,int x,int y)
44 {
45     if (l==x&&r==y) return fuck[p];
46     else
47     {
48         int mid=(l+r)>>1;
49         if (y<=mid) return query_fuck(l,mid,p*2,x,y);
50         else if(x>=mid+1) return query_fuck(mid+1,r,p*2+1,x,y);
51         else return min(query_fuck(l,mid,p*2,x,mid),query_fuck(mid+1,r,p*2+1,mid+1,y));
52     }
53 }
54 int main()
55 {
56     scanf("%d%d",&n,&q);
57     for (int i=1;i<=n;i++)
58         scanf("%d",&a[i]);
59     build(1,n,1);
60     build_fuck(1,n,1);
61     while (q--)
62     {
63         scanf("%d%d",&x,&y);
64         printf("%d
",query(1,n,1,x,y)-query_fuck(1,n,1,x,y));
65     }
66 }