CODEVS3304 fruit sister visits Fruit Street

(I wanted to finish writing about the fruit sister series, but I suddenly found that I will leave noip tomorrow, so ...) Description Description

The fruit sister was in a good mood today and came to the fruit Street.

There are n fruit shops in Fruit Street, in a straight line structure, numbered 1 ~ N. Each store can buy or sell fruit, and the same store can sell at the same price.

The fruit sister who has learned oi quickly found a way to make money: buy a fruit in a fruit shop and then sell it to another store to make a difference.

Just when the fruit sister had a crush, cgh suddenly appeared. In order to embarrass the fruit sister, he gave m problems. Each problem required the fruit sister to depart from the x store to the y store, on the way, you can only choose one store to buy one fruit, and then choose one store (which can be the same store, but cannot go back) to sell it. Find the maximum amount of money you can earn for each problem.

 

Idea: The linear problem of the fruit sister series is maintained using the line segment tree, saving time. Use the four items in the Struct (maximum range, small value, maximum range trade value (positive and inverse )). Here is a clever thing, that is, for the range [a, B], its maximum and small values are relatively simple, however, the maximum trade value is updated based on the difference between the maximum trade value of the left and right child ranges and the maximum and minimum values of the left and right ranges. I started to use four arrays, and later found tle, so I learned from the dada program and changed it to a struct. As a result, I made another magic mistake (directly changed the value of the tree ), you only need to return the value...

 

# Include <iostream> # include <cstdio> using namespace std; struct use {int maxi, mini, zan, fan;} tr [800000]; int a [200001] = {0}, n; void build (int I, int l, int r) {int mid; if (l = r) {tr [I]. maxi = tr [I]. mini = a [l]; return;} mid = (l + r)/2; build (I * 2, l, mid); build (I * 2 + 1, mid + 1, r); tr [I]. maxi = max (tr [I * 2]. maxi, tr [I * 2 + 1]. maxi); tr [I]. mini = min (tr [I * 2]. mini, tr [I * 2 + 1]. mini); tr [I]. zan = max (tr [I * 2]. zan, max (tr [I * 2 + 1]. zan, tr [I * 2 + 1]. maxi-tr [I * 2]. mini); tr [I]. fan = max (tr [I * 2]. fan, max (tr [I * 2 + 1]. fan, tr [I * 2]. maxi-tr [I * 2 + 1]. mini);} struct use work (int I, int l, int r, int x, int y) {int mid, maxn = 0; struct use tem, temm, temi; if (x> r | y <l) return (use) {-1, 2100000000,-1,-1}; if (x <= l & r <= y) return tr [I]; mid = (l + r)/2; tem = work (I * 2, l, mid, x, y ); temm = work (I * 2 + 1, mid + 1, r, x, y); temi. maxi = max (tem. maxi, temm. maxi); temi. mini = min (tem. mini, temm. mini); temi. zan = max (tem. zan, max (temm. zan, temm. maxi-tem.mini); temi. fan = max (tem. fan, max (temm. fan, tem. maxi-temm.mini); return temi;} int main () {int m, I, j, x, y; struct use ans; scanf ("% d", & n ); for (I = 1; I <= n; ++ I) scanf ("% d", & a [I]); build (1, 1, n ); scanf ("% d", & m); for (I = 1; I <= m; ++ I) {scanf ("% d", & x, & y); if (x = y) printf ("0 \ n"); else {if (x <y) {ans = work (1, 1, n, x, y); printf ("% d \ n", ans. zan);} else {ans = work (1, 1, n, y, x); printf ("% d \ n", ans. fan );}}}}

RZUC Code

 

CODEVS3304 fruit sister visits Fruit Street