Vasya got really tired of these credits (from problem F) and now wants to earn the money himself! He decided to make a contest to gain a profit.
Vasya has n problems to choose from. They are numbered from 1 to n. The difficulty of the i-th problem is di. Moreover, the problems are given in the increasing order by their difficulties. The difficulties of all tasks are pairwise distinct. In order to add the i-th problem to the contest you need to pay ci burles to its author. For each problem in the contest Vasya gets a burles.
In order to create a contest he needs to choose a consecutive subsegment of tasks.
So the total earnings for the contest are calculated as follows:
if Vasya takes problem i to the contest, he needs to pay ci to its author;
for each problem in the contest Vasya gets a burles;
let gap(l,r)=maxl≤i<r(di+1−di)2. If Vasya takes all the tasks with indices from l to r to the contest, he also needs to pay gap(l,r). If l=r then gap(l,r)=0.
Calculate the maximum profit that Vasya can earn by taking a consecutive segment of tasks.
Input
The first line contains two integers n and a (1≤n≤3⋅105, 1≤a≤109) — the number of proposed tasks and the profit for a single problem, respectively.
Each of the next n lines contains two integers di and ci (1≤di,ci≤109,di<di+1).
Output
Print one integer — maximum amount of burles Vasya can earn.
Examples
inputCopy
5 10
1 15
5 3
6 11
7 2
11 22
outputCopy
13
inputCopy
3 5
1 8
2 19
3 11
outputCopy
0
题意：
给你n种题目，然后给你他们的难度di和花费ci，你能出一段区间的题目，出一道题目可以得到a元，但是要花费区间所有题目的价格，还要花费在这个区间里，相邻两个题目难度差值的平方的最大值，问你最后可以赚最多多少钱。
题解：
区间的变化，a的值和ci的和都是可以预处理出来的，那么我们需要考虑的就是d，我们枚举每一位上的d，记录它最远能够管辖的区间，sta存放的是到当前位置为止，有哪些开头可以管辖到当前位置，和他们最早能管辖到的位置的c的前缀和。举个例子：
有这三个位置，他们一定是单调递减的，因为如果有递增的话，前面那个小的d会被后面大的d取代掉，所以就变成这样。sta记录的就是10的位置和蓝色的线，5的位置和蓝色的线。
那如果现在来了一个6，他就能管辖到5所管辖到的位置，而且这就是为什么要用LIFO的原因。
之后我们还需要有一个东西可以自动排序蓝色线的最小值，在5被删掉的时候，蓝色的线也可以被删掉，因为它无法管辖到下一个位置了。那就用multiset。
其实也是看别人的代码想出来的
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pa pair<int,ll>
const int N=3e5+5;
ll d[N],c[N];
ll dis(int x)
{
return (d[x+1]-d[x])*(d[x+1]-d[x]);
}
int main()
{
int n;
ll a,ans=0;
scanf("%d%lld",&n,&a);
for(int i=1;i<=n;i++)
{
scanf("%lld%lld",&d[i],&c[i]);
ans=max(ans,a-c[i]);
c[i]=c[i-1]+a-c[i];
}
stack<pa>st;
multiset<ll>ms;
for(int i=1;i<=n;i++)
{
if(!ms.empty())
ans=max(ans,c[i]-*ms.begin());
ll dist=dis(i),pre=c[i-1];
while(!st.empty()&&dis(st.top().first)<=dist)
{
pre=min(pre,st.top().second);
ms.erase(ms.find(st.top().second+dis(st.top().first)));
st.pop();
}
st.push({i,pre});
ms.insert(dist+pre);
}
printf("%lld\n",ans);
return 0;
}