bzoj 1636: [Usaco2007 Jan]Balanced Lineup -- 线段树

1636: [Usaco2007 Jan]Balanced Lineup

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 772  Solved: 560线段树裸题。。。

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.

每天,农夫 John 的N(1 <= N <= 50,000)头牛总是按同一序列排队. 有一天, John 决定让一些牛们玩一场飞盘比赛. 他准备找一群在对列中为置连续的牛来进行比赛. 但是为了避免水平悬殊,牛的身高不应该相差太大. 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 <= 身高 <= 1,000,000). 他想知道每一组里面最高和最低的牛的身高差别.

注意: 在最大数据上, 输入和输出将占用大部分运行时间. 

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.

第1行：N,Q

第2到N+1行：每头牛的身高

第N+2到N+Q+1行：两个整数A和B，表示从A到B的所有牛。（1<=A<=B<=N）

Output

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.

 

输出每行一个数，为最大数与最小数的差

Sample Output

6
3
0

 

#include<cstdio>
#include <iostream>
#define M 50010
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
struct tree{int l,r,minn,maxx;}tr[*M];
int a[M];
void make(int l,int r,int p)
{
    tr[p].l=l;
    tr[p].r=r;
    if(l==r){
        tr[p].minn=a[l];
        tr[p].maxx=a[l];
        return ;
    }
    int mid=(l+r)>>;
    make(l,mid,p<<);
    make(mid+,r,p<<|);
    tr[p].minn=min(tr[p<<].minn,tr[p<<|].minn);
    tr[p].maxx=max(tr[p<<].maxx,tr[p<<|].maxx);
}
int fmin(int l,int r,int x)
{
    if(tr[x].l==l&&tr[x].r==r) return tr[x].minn;
    int mid=(tr[x].l+tr[x].r)>>,q=x<<;
    if(r<=mid) return fmin(l,r,q);
    else if(l>mid) return fmin(l,r,q+);
    else return min(fmin(l,mid,q),fmin(mid+,r,q+));
}
int fmax(int l,int r,int x)
{
    if(tr[x].l==l&&tr[x].r==r) return tr[x].maxx;
    int mid=(tr[x].l+tr[x].r)>>;
    if(r<=mid) return fmax(l,r,x<<);
    else if(l>mid) return fmax(l,r,x<<|);
    else return max(fmax(l,mid,x<<),fmax(mid+,r,x<<|));
}
int main()
{
    int n,m,i,x,y;
    scanf("%d%d",&n,&m);
    for(i=;i<=n;i++) scanf("%d",&a[i]);
    make(,n,);
    for(i=;i<m;i++){
        scanf("%d%d",&x,&y);
        printf("%d\n",fmax(x,y,)-fmin(x,y,));
    }
}