POJ 3264 Balanced Lineup

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题意：给出初始化的区间值，m次查询。每次查询区间[a，b]的最大值－最小值

分析：线段树更新: 无更新查询:区间查询，建立线段树的时候，每个结点存储左右子树的最大值和最小值，查询时直接访问区间最大值和最小值，不需要查找到最低，查询时间复杂度O(logN)

//2384K1594MS#include <cstdio>#include <cmath>#include <iostream>#include <climits>#define MAXN 50010#define L(X) ((X)<<1)#define R(X) (((X)<<1) | 1 )#define MID(X,Y) ( (X+Y) >> 1 )#define MAX(X,Y) ( (X)>(Y) ? (X) : (Y) )#define MIN(X,Y) ( (X)<(Y) ? (X) : (Y) )struct node{    int left , right ;    int min , max ;}node[MAXN*4];int num[MAXN] ;int ans_min , ans_max ;void Build_Tree ( int const left , int const right , int const n ){    node[n].left = left ;    node[n].right = right ;    if ( left < right )    {        int mid ;        mid = MID(left,right) ;        Build_Tree ( left , mid , L(n)) ;        Build_Tree ( mid + 1 , right , R(n)) ;    }    else if ( left == right )    {        node[n].min = node[n].max =  num[left] ;        return ;    }    node[n].min = MIN ( node[L(n)].min , node[R(n)].min ) ;    node[n].max = MAX ( node[L(n)].max , node[R(n)].max ) ;}void Query ( int const left , int const right , int const n ){    if ( left == node[n].left && right == node[n].right )    {        ans_max = MAX ( node[n].max , ans_max ) ;        ans_min = MIN ( node[n].min , ans_min ) ;        return ;    }    int mid ;    mid = MID ( node[n].left , node[n].right ) ; //注意，求的中值不是(left + right)/2，注意区间范围    if ( left > mid )    {        Query ( left , right , R(n) ) ;    }    else  if ( right <= mid )    {        Query ( left , right , L(n) ) ;    }    else    {        Query ( left , mid , L(n) ) ;        Query ( mid + 1 , right , R(n) ) ;    }}intmain ( ){    int n , m ;    scanf ("%d%d" , & n , & m ) ;    for ( int i = 1 ; i <= n ; i ++ )    {        scanf ("%d" , & num[i] ) ;    }    Build_Tree ( 1 , n , 1 ) ;    for ( int i = 1 ; i <= m ; i ++ )    {        ans_min = INT_MAX ;        ans_max = 0 ;        int left , right ;        scanf ("%d%d" , & left , & right ) ;        Query ( left , right , 1 ) ;        printf ("%d\n" , ans_max - ans_min ) ;    }    return 0 ;}