POJ 3264 Balanced Lineup

题目传送门

树状数组求最大最小值，一次成型，不能跑两回啊！

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>
#include <math.h>
#include <cstdio>
using namespace std;
const int N = 50010;
int t1[N], t2[N];
int a[N];
int n, q;

int lowbit(int x) {
    return x & -x;
}

void update(int x, int v) {
    while (x <= n) {
        t1[x] = max(t1[x], v);
        t2[x] = min(t2[x], v);
        x += lowbit(x);
    }
}

int query(int x, int y) {
    int m1 = a[x], m2 = a[x];
    while (true) {
        m1 = max(m1, a[y]), m2 = min(m2, a[y]);
        if (x == y) break;
        for (y -= 1; y - x >= lowbit(y); y -= lowbit(y))
            m1 = max(m1, t1[y]), m2 = min(m2, t2[y]);
    }
    return m1 - m2;
}

int main() {
    memset(t2, 0x3f, sizeof(t2));
    scanf("%d%d", &n, &q);

    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        update(i, a[i]);
    }

    while (q--) {
        int a, b;
        scanf("%d%d", &a, &b);
        printf("%d\n", query(a, b));
    }
    return 0;
}

标签：include,int,lowbit,t2,POJ,m2,Balanced,m1,Lineup
来源： https://www.cnblogs.com/littlehb/p/16224395.html