[单调栈]Cow Acrobats

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

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.

题目大意

求出每一个元素右边第一个比它大的元素（仰望对象）

思路分析

朴素算法就是遍历一遍整个序列，对于每个元素，依次往右扫，扫到第一个比他大的元素，就记录下来。这样的时间复杂度为O（n^2）,肯定要超时的。我们发现每次往右扫有重复的过程，那么能不能把每次扫描的过程记录下来呢？
首先肯定扫的话是从右往左扫，最右边的元素没有仰望对象，记成0，倒数第二个元素仰望对象肯定是最后边的元素，依次往左遍历，如果新加入的数字比之前的数字还要大的话，要之前的数字也就没有用了（对于这个数来说，我要找比它大的数，比它小的数没用，对于这个数左边的数，这个数都已经比那些数要大，跟他们就更没关系了），就让他们出栈，而出栈后栈顶元素就是我们要找的栈顶元素。