Farmer John’s N (1 <= N <= 50,000) cows (numbered 1…N) are planning to run away and join the circus. Their hoofed feet prevent them from tightrope walking and swinging from the trapeze (and their last attempt at firing a cow out of a cannon met with a dismal failure). Thus, they have decided to practice performing acrobatic stunts.
The cows aren’t terribly creative and have only come up with one acrobatic stunt: standing on top of each other to form a vertical stack of some height. The cows are trying to figure out the order in which they should arrange themselves ithin this stack.
Each of the N cows has an associated weight (1 <= W_i <= 10,000) and strength (1 <= S_i <= 1,000,000,000). The risk of a cow collapsing is equal to the combined weight of all cows on top of her (not including her own weight, of course) minus her strength (so that a stronger cow has a lower risk). Your task is to determine an ordering of the cows that minimizes the greatest risk of collapse for any of the cows.
Input
-
Line 1: A single line with the integer N.
-
Lines 2…N+1: Line i+1 describes cow i with two space-separated integers, W_i and S_i.
Output
- Line 1: A single integer, giving the largest risk of all the cows in any optimal ordering that minimizes the risk.
Sample Input
3
10 3
2 5
3 3
Sample Output
2
Hint
OUTPUT DETAILS:
Put the cow with weight 10 on the bottom. She will carry the other two cows, so the risk of her collapsing is 2+3-3=2. The other cows have lower risk of collapsing.
題目大意
求出每一個元素右邊第一個比它大的元素（仰望對象）
思路分析
樸素算法就是遍歷一遍整個序列，對於每個元素，依次往右掃，掃到第一個比他大的元素，就記錄下來。這樣的時間複雜度爲O（n^2）,肯定要超時的。我們發現每次往右掃有重複的過程，那麼能不能把每次掃描的過程記錄下來呢？
首先肯定掃的話是從右往左掃，最右邊的元素沒有仰望對象，記成0，倒數第二個元素仰望對象肯定是最後邊的元素，依次往左遍歷，如果新加入的數字比之前的數字還要大的話，要之前的數字也就沒有用了（對於這個數來說，我要找比它大的數，比它小的數沒用，對於這個數左邊的數，這個數都已經比那些數要大，跟他們就更沒關係了），就讓他們出棧，而出棧後棧頂元素就是我們要找的棧頂元素。