poj3264-Balanced Lineup(点更新，区间最值)

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题意：

给出每个位置奶牛高度，求出l,r这个区间段最高牛和最低的牛的差值

思路：

线段树维护最值

#include <stdio.h>#include <algorithm>#include <cstring>#include <iostream>using namespace std;const int maxn= 50005;const int inf=0x3f3f3f3f;int a[maxn];struct node{    int left,right,maxx,minn,num;}tree[4*maxn];void Build(int i,int left,int right){    tree[i].left=left,tree[i].right=right;    if(left==right)    {        tree[i].num=a[ tree[i].left ];        tree[i].minn=tree[i].num;        tree[i].maxx=tree[i].num;        return ;    }    int mid=(tree[i].left+tree[i].right)>>1;    Build(i<<1,left,mid);    Build(i<<1|1,mid+1,right);    tree[i].minn=min(tree[i<<1].minn,tree[i<<1|1].minn);    tree[i].maxx=max(tree[i<<1].maxx,tree[i<<1|1].maxx);    return ;}int ansmax,ansmin;void query(int i,int left,int right){    if(tree[i].right==right&&tree[i].left==left)    {        ansmax=max(ansmax,tree[i].maxx);        ansmin=min(ansmin,tree[i].minn);        return ;    }    int mid=(tree[i].left+tree[i].right)>>1;    if(left>mid)    {        query(i<<1|1,left,right);    }    else if(right<=mid)    {        query(i<<1,left,right);    }    else    {        query(i<<1,left,mid);        query(i<<1|1,mid+1,right);    }    return ;}int main(){    int n,q;    scanf("%d%d",&n,&q);    for(int i=1;i<=n;i++)        scanf("%d",&a[i]);    Build(1,1,n);    while(q--)    {        int l,r;        scanf("%d%d",&l,&r);        ansmax=0,ansmin=inf;        query(1,l,r);        printf("%d\n",ansmax-ansmin);    }}

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