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.
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
6 3
1
7
3
4
2
5
1 5
4 6
2 2
Sample Output
6
3
0
Source
USACO 2007 January Silver
View Code
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 }