POJ3438 Balanced Lineup 基本线段树

线段树的模板题，另外此题还能用SparseTable解。

#include <stdio.h>struct CowRange{    int l,u;    CowRange *left,*right;    int max,min;};CowRange* BuildInTree(int l, int u, int *height){    CowRange *root = new CowRange();    root->l = l;root->u = u;    if(l!=u){    //递归构建左右子树，并初始化max和min值        root->left = BuildInTree(l,(l+u)/2,height);        root->right = BuildInTree((l+u)/2+1,u,height);        root->max = root->left->max > root->right->max?root->left->max:root->right->max;        root->min = root->left->min < root->right->min?root->left->min:root->right->min;    }    else{    //只有一个元素，初始化max和min值，也可以把left和right指针置为null        root->max = height[l];        root->min = height[l];    }    return root;}void Query(CowRange *root, int l, int u, int &max, int &min){    if(l <= root->l && root->u<=u){    //只有当[l,u]包含root表示的区间时，root的max和min值才有效        if(max < root->max)max = root->max;        if(min > root->min)min = root->min;    }    else{        if(l <= (root->l+root->u)/2){//l在左子树中            Query(root->left,l,u,max,min);        }        if(u >= (root->l+root->u)/2+1){//u在右子树中            Query(root->right,l,u,max,min);        }    }}int main(){    int N,Q;    scanf("%d %d",&N,&Q);    int *height = new int[N+1];    for(int i=1;i<=N;i++)scanf("%d",height+i);    CowRange *root = BuildInTree(1,N,height);    int l,u;    int max,min;    for(int i=0;i<Q;i++){        scanf("%d %d",&l,&u);        max = min = height[l];        Query(root,l,u,max,min);        printf("%d\n",max-min);    }    return 0;}

最后结果，如果用cin，cout会超时。5012kB780ms1657 B
/*
3438:Balanced Lineup
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时间限制: 5000ms 内存限制: 65536kB
描述
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.


输入
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.
输出
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.
样例输入
6 3
1
7
3
4
2
5
1 5
4 6
2 2
样例输出
6
3
0
*/