18.1.5 POJ 3264 Balanced Lineup

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

 1 #include <iostream>
 2 #include <string.h>
 3 #include <algorithm>
 4 #include <stack>
 5 #include <string>
 6 #include <math.h>
 7 #include <queue>
 8 #include <stdio.h>
 9 #include <string.h>
10 #include <set>
11 #include <vector>
12 #define maxn 50005
13 #define inf 999999
14 #define EPS 1e-10
15 #define lowbit(x) (int)(x&(-x))
16 using namespace std;
17 
18 struct node {
19     int l, r;
20     int max, min;
21 }tree[maxn<<2];
22 int n, m;
23 int a[maxn];
24 
25 void build(int rt, int l, int r) {
26     int mid = (l + r) / 2;
27     tree[rt].l = l, tree[rt].r = r;
28     if (l == r) {
29         tree[rt].max = a[l];
30         tree[rt].min = a[l];
31         return;
32     }
33     build(2 * rt, l, mid), build(2 * rt + 1, mid + 1, r);
34     int maxv = max(tree[2 * rt].max, tree[2 * rt + 1].max);
35     int minv = min(tree[2 * rt].min, tree[2 * rt + 1].min);
36     tree[rt].max = maxv, tree[rt].min = minv;
37 }
38 
39 pair<int,int> query(int rt, int l, int r) {
40     int ll = tree[rt].l, rr = tree[rt].r;
41     int mid = (ll + rr) / 2;
42     if (ll == l && rr == r) 
43         return pair<int,int>(tree[rt].max,tree[rt].min);
44     if (r <= mid)
45         return query(rt * 2, l, r);
46     else if (l > mid)
47         return query(rt * 2 + 1, l, r);
48     pair<int, int> ret1=query(rt*2,l,mid), ret2=query(rt*2+1,mid+1,r);
49     int maxv = max(ret1.first, ret2.first);
50     int minv = min(ret1.second, ret2.second);
51     return pair<int, int>(maxv, minv);
52 }
53 
54 void init() {
55     scanf("%d%d", &n, &m);
56     for (int i = 1; i <= n; i++)
57         scanf("%d", &a[i]);
58     build(1, 1, n);
59     for (int i = 1; i <= m; i++) {
60         int x, y;
61         scanf("%d%d", &x, &y);
62         pair<int, int>ans = query(1, x, y);
63         printf("%d
", ans.first - ans.second);
64     }
65 }
66 
67 int main() {
68     init();
69     return 0;
70 }

View Code

注定失败的战争，也要拼尽全力去打赢它；就算输，也要输得足够漂亮。