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| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 38139 | Accepted: 17863 | |
| Case Time Limit: 2000MS | ||
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.
6 3
1
7
3
4
2
5
1 5
4 6
2 2
6
3
0
随手打了个线段树，竟然A了，但是慢的一比、、、、
CODE:
#include <iostream> #include <cstdio> #include <cstring> #define REP(i, s, n) for(int i = s; i <= n; i ++) #define REP_(i, s, n) for(int i = n; i >= s; i --) #define max(a, b) a > b ? a : b #define min(a, b) a < b ? a : b #define MAX_N 50000 + 10 using namespace std; int n, q, data[MAX_N]; struct node{ int l, r, mx, mn; }a[MAX_N << 2]; void update(int i){ int t1 = i << 1, t2 = t1 + 1; a[i].mx = max(a[t1].mx, a[t2].mx); a[i].mn = min(a[t1].mn, a[t2].mn); } void maketree(int i, int l, int r){ a[i].l = l; a[i].r = r; if(a[i].l == a[i].r){ a[i].mx = data[l]; a[i].mn = data[l]; return; } int mid = (a[i].l + a[i].r) >> 1, t1 = i << 1, t2 = t1 + 1; maketree(t1, l, mid); maketree(t2, mid + 1, r); update(i); } int Query_Max(int i, int l, int r){ if(a[i].l == l && a[i].r == r) return a[i].mx; int mid = (a[i].l + a[i].r) >> 1, t1 = i << 1, t2 = t1 + 1; if(r <= mid) return Query_Max(t1, l, r); else if(l > mid) return Query_Max(t2, l, r); else return max(Query_Max(t1, l, mid), Query_Max(t2, mid + 1, r)); } int Query_Min(int i, int l, int r){ if(a[i].l == l && a[i].r == r) return a[i].mn; int mid = (a[i].l + a[i].r) >> 1, t1 = i << 1, t2 = t1 + 1; if(r <= mid) return Query_Min(t1, l, r); else if(l > mid) return Query_Min(t2, l, r); else return min(Query_Min(t1, l, mid), Query_Min(t2, mid + 1, r)); } int main(){ scanf("%d%d", &n, &q); REP(i, 1, n) scanf("%d", &data[i]); maketree(1, 1, n); int x, y; while(q --){ scanf("%d%d", &x, &y); printf("%d\n", Query_Max(1, x, y) - Query_Min(1, x, y)); } return 0; }
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原文地址：http://www.cnblogs.com/ALXPCUN/p/4550786.html
