Balanced Lineup（线段树——根据区间找最值）

Balanced Lineup

Time Limit: 5000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld      Java class name:Main

SubmitStatus PID: 3383

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 andQ.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cowi
Lines N+2..N+Q+1: Two integers A and B (1 ≤A ≤ B ≤ N), representing the range of cows from A toB 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

6 31734251 54 62 2

Sample Output

630

   query（）函数

#include <stdio.h>#include <algorithm>#define MAX 50000using namespace std;int a[MAX+10];int low, hei;struct node{    int mx, mn;}seg[MAX*4+10];void build(int node, int l, int r){    if(l == r){        seg[node].mn = seg[node].mx = a[l];        return;    }    int m = (l + r) / 2;    build(node * 2, l, m);    build(node * 2 + 1, m + 1, r);    seg[node].mn = seg[node*2].mn < seg[node*2+1].mn ? seg[node*2].mn : seg[node*2+1].mn;    seg[node].mx = seg[node*2].mx > seg[node*2+1].mx ? seg[node*2].mx : seg[node*2+1].mx;}void query (int node, int l, int r, int s, int e){    if (l == s && r == e)    {        low = min (low, seg[node].mn);        hei = max (hei, seg[node].mx);        return ;    }    int m = (l + r) / 2;    if (e <= m)        query (2 * node, l, m, s, e);    else if (s >= m + 1)        query (2 * node + 1, m + 1, r, s, e);    else    {        query(node * 2, l, m, s, m);        query (2 * node + 1, m + 1, r, m + 1, e);    }}int main(){    int n, m;    scanf("%d %d", &n, &m);    for(int i = 1; i <= n; i ++ ){        scanf("%d", &a[i]);    }    build(1, 1, n);    while(m -- ){        int a, b;        scanf("%d %d", &a, &b);        low = 1000001;        hei = 0;        query(1, 1, n, a, b);        printf("%d\n", hei - low);    }    return 0;}