# [BZOJ] 1636: [Usaco2007 Jan]Balanced Lineup

## 1636: [Usaco2007 Jan]Balanced Lineup

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 909  Solved: 644
[Submit][Status][Discuss]

## 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.

John 准备了Q (1 <= Q <= 180,000) 个可能的牛的选择和所有牛的身高 (1 <=

## 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.

6 3
1
7
3
4
2
5
1 5
4 6
2 2

## Sample Input

* 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.

6
3
0

Silver

## Code  1 #include<cstdio>
2 #include<iostream>
3 #define mid (L+R)/2
4 #define lc (rt<<1)
5 #define rc (rt<<1|1)
6 #define maxn 50000
7 using namespace std;
8
9 const int inf = 0x3f3f3f3f;
10
11 int n,q,a,b;
12
13 struct node{
14     int maxx,minn;
15 }Tree[maxn*10];
16
17 void maintain(int rt){
18     Tree[rt].maxx = max(Tree[lc].maxx,Tree[rc].maxx);
19     Tree[rt].minn = min(Tree[lc].minn,Tree[rc].minn);
20 }
21
22 void build(int rt,int L,int R){
23     if(L == R){
24         scanf("%d",&Tree[rt].maxx);
25         Tree[rt].minn = Tree[rt].maxx;
26     }else{
27         build(lc,L,mid);
28         build(rc,mid+1,R);
29
30         maintain(rt);
31     }
32 }
33
34 int max_query(int rt,int L,int R,int qL,int qR){
35     if(qL <= L && R <= qR){
36         return Tree[rt].maxx;
37     }else{
38         int ans = 0;
39         if(qL <= mid) ans = max(ans,max_query(lc,L,mid,qL,qR));
40         if(qR > mid) ans = max(ans,max_query(rc,mid+1,R,qL,qR));
41         return ans;
42     }
43 }
44
45 int min_query(int rt,int L,int R,int qL,int qR){
46     if(qL <= L && R <= qR){
47         return Tree[rt].minn;
48     }else{
49         int ans = inf;
50         if(qL <= mid) ans = min(ans,min_query(lc,L,mid,qL,qR));
51         if(qR > mid) ans = min(ans,min_query(rc,mid+1,R,qL,qR));
52         return ans;
53     }
54 }
55
56 int main(){
57     scanf("%d%d",&n,&q);
58     build(1,1,n);
59
60     for(int i = 1;i <= q;i++){
61         scanf("%d%d",&a,&b);
62         printf("%d\n",max_query(1,1,n,a,b)-min_query(1,1,n,a,b));
63     }
64
65
66     return 0;
67 }
qwq

posted @ 2017-09-02 21:15  Leviaton  阅读(165)  评论(0编辑  收藏  举报