[线段树单点更新区间查询多个值] G - Balanced Lineup POJ - 3264

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解决一个关于从连续范围内选择高度相近的牛参加游戏的问题，通过构建线段树来快速查询区间内的最大值和最小值。

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 61985		Accepted: 28950
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

Sample Output

6
3
0

Source

USACO 2007 January Silver

#include <iostream>
#include <cstdio>
using namespace std;

const int mn = 50010;
const int mq = 200010;
const int inf = 0x7fffffff;

struct node
{
    int l, r, minn, maxx;
} t[4 * mn];

void build(int i, int l, int r)
{
    t[i].l = l, t[i].r = r;
    if (l == r)
    {
        int a;
        scanf("%d", &a);
        t[i].maxx = t[i].minn = a;
        return;
    }
    int m = (l + r) / 2;
    build(2 * i, l, m);
    build(2 * i + 1, m + 1, r);
    t[i].minn = min(t[2 * i].minn, t[2 * i + 1].minn);
    t[i].maxx = max(t[2 * i].maxx, t[2 * i + 1].maxx);
    return;
}

int x, y = inf;     /// 全局变量记录区间最值
void query(int i, int a, int b)
{
    int l = t[i].l, r = t[i].r;
    if (a == l && r == b)
    {
        x = max(x, t[i].maxx);  /// 区间包含 更新最值
        y = min(y, t[i].minn);
        return;
    }
    
    int m = (l + r) / 2;
    if (b <= m)
        query(2 * i, a, b);
    else if (a > m)
        query(2 * i + 1, a, b);
    else    /// 分别查询并更新
    {
        query(2 * i, a, m);
        query(2 * i + 1, m + 1, b);
    }
    return;
}
int main()
{
    int n, q;
    scanf("%d %d", &n, &q);
    build(1, 1, n);
    while (q--)
    {
        int a, b;
        scanf("%d %d", &a, &b);
        
        x = 0, y = inf;     /// 初始化
        query(1, a, b);
        printf("%d\n", x - y);
    }
    return 0;
}