BZOJ-1699 Balanced Lineup 线段树区间最大差值

Balanced Lineup
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 41548 Accepted: 19514
Case Time Limit: 2000MS

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

Source
USACO 2007 January Silver

题目大意：农夫JOHN让我们求一个区间上的最大差值（最大值-最小值）

维护两棵线段树即可，一棵max，一棵min，然后求差值，自己第一次摸索写多棵线段树姿势，感觉写的还可以，以后待改进...

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 50001
int maxs[maxn<<2]={0},mins[maxn<<2]={0};

void updata1(int now)
{
    maxs[now]=max(maxs[now<<1],maxs[now<<1|1]); 
}

void updata2(int now)
{
    mins[now]=min(mins[now<<1],mins[now<<1|1]);
}

void point_change(int l,int r,int now,int loc,int num)
{
    if (l==r)
        {
            maxs[now]=num;
            mins[now]=num;
            return;
        }
    int mid=(l+r)>>1;
    if (loc<=mid)
        point_change(l,mid,now<<1,loc,num);
    else
        point_change(mid+1,r,now<<1|1,loc,num);
    updata1(now);
    updata2(now);
}

int query(int L,int R,int now,int l,int r,bool f)//f为true为求最大，false为求最小 
{
    if (L<=l && R>=r)
        if (f==true)
            return maxs[now];
        else
            return mins[now];
    int mid=(l+r)>>1;
    if (f==true)
        {
            int ans=0;
            if (L<=mid)
                ans=max(ans,query(L,R,now<<1,l,mid,true));
            if (R>mid)
                ans=max(ans,query(L,R,now<<1|1,mid+1,r,true));
            return ans;
        }
    else
        {
            int ans=100000000;
            if (L<=mid)
                ans=min(ans,query(L,R,now<<1,l,mid,false));
            if (R>mid)
                ans=min(ans,query(L,R,now<<1|1,mid+1,r,false));
            return ans;         
        }   
}

int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    //build(1,n,1);
    for (int i=1; i<=n; i++)
        {
            int x;
            scanf("%d",&x);
            point_change(1,n,1,i,x);
        }
    for (int i=1; i<=m; i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            int answer=query(x,y,1,1,n,true)-query(x,y,1,1,n,false);
            printf("%d\n",answer);
        }
    return 0;
}