codeforces 436F Banners

http://www.elijahqi.win/2018/03/07/codeforces-436f-banners/
All modern mobile applications are divided into free and paid. Even a single application developers often release two versions: a paid version without ads and a free version with ads. Suppose that a paid version of the app costs p p ( p p is an integer) rubles, and the free version of the application contains c c ad banners. Each user can be described by two integers: a_{i} a i ​ — the number of rubles this user is willing to pay for the paid version of the application, and b_{i} b i ​ — the number of banners he is willing to tolerate in the free version. The behavior of each member shall be considered strictly deterministic: if for user i i , value b_{i} b i ​ is at least c c , then he uses the free version, otherwise, if value a_{i} a i ​ is at least p p , then he buys the paid version without advertising, otherwise the user simply does not use the application. Each user of the free version brings the profit of c×w c×w rubles. Each user of the paid version brings the profit of p p rubles. Your task is to help the application developers to select the optimal parameters p p and c c . Namely, knowing all the characteristics of users, for each value of c c from 0 0 to $(max b_{i})+1$ you need to determine the maximum profit from the application and the corresponding parameter p p . 输入输出格式 输入格式： The first line contains two integers n n and w w $(1<=n<=10^{5}; 1<=w<=10^{5})$ — the number of users and the profit from a single banner. Each of the next n n lines contains two integers a_{i} a i ​ and b_{i} b i ​ (0<=a_{i},b_{i}<=10^{5}) (0<=a i ​ ,b i ​ <=10 5 ) — the characteristics of the i i -th user. 输出格式： Print $(max b_{i})+2$ lines, in the i i -th line print two integers: pay pay — the maximum gained profit at c=i-1 c=i−1 , p p (0<=p<=10^{9}) (0<=p<=10 9 ) — the corresponding optimal app cost. If there are multiple optimal solutions, print any of them. 输入输出样例 输入样例#1： 复制 2 1 2 0 0 2 输出样例#1： 复制 0 3 3 2 4 2 2 2 输入样例#2： 复制 3 1 3 1 2 2 1 3 输出样例#2： 复制 0 4 3 4 7 3 7 2 4 2
给定每个客人忍受广告的篇幅数 如果没超过了这个值 他会使用免费版本 再给定一个
付费值 如果不到他这个承受能力则他会使用付费版 那么首先针对这个b对所有的客人排序 这样的话每次询问c的时候广告的收益可以o(1)算 但是这个付费的价值我们就不能确定了，因为随着广告的递增我会有人去用付费版本那么我们针对付费分块 每个块里维护一个凸包 维护付费价格在这个区间的最优答案的下凸曲线 设i＜j d(i)＞d(j) 那么当我们增加这个付费的人数为x的时候如果出现d(i)+x*i≤d(j)+x*j那么显然我i再也不可能是答案了，所以用d(i)-d(j)/(j-i)维护这个斜率单增的下凸曲线即可 复杂度为什么对得，因为整块的直接打标记即可 如果不是整块也只需要暴力重构sqrt(n)大小的块即可

#include<cmath>
#include<cstdio>
#include<algorithm>
#define N 110000
#define N1 350
#define ll long long
using namespace std;
inline char gc(){
    static char now[1<<16],*S,*T;
    if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0,f=1;char ch=gc();
    while(ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=gc();}
    while(ch<='9'&&ch>='0') x=x*10+ch-'0',ch=gc();
    return x*f;
}
struct node{
    int a,b;
    inline friend bool operator <(const node &a,const node &b){
        return a.b<b.b;
    }
}data[N];
int q[N1][N1],ql[N1],qr[N1],left[N1],right[N1],cnt[N1],b[N],n,w,nn;
ll d[N];
inline double slope(int i,int j){return (double)(d[i]-d[j])/(j-i);}
inline void solve(int id,int x){
    while(ql[id]<qr[id]&&slope(q[id][ql[id]],q[id][ql[id]+1])<=x) ++ql[id];
}
inline void rebuild(int x,int l,int r){
    ql[x]=1,qr[x]=0;
    for (int i=l;i<=r;++i){
        while(ql[x]<qr[x]&&slope(q[x][qr[x]],i)<slope(q[x][qr[x]-1],q[x][qr[x]])) --qr[x];
        q[x][++qr[x]]=i;
    }solve(x,0);
}
inline void add(int x){
    if (!x) return;int id=b[x];
    for (int i=1;i<id;++i) solve(i,++cnt[i]);
    for (int i=left[id];i<=right[id];++i) d[i]+=(ll)cnt[id]*i;
    for (int i=left[id];i<=x;++i) d[i]+=i;cnt[id]=0;rebuild(id,left[id],right[id]);
}
int main(){
//  freopen("cf436f.in","r",stdin);
    n=read();w=read();int mx1=0,mx2=0,x1,x2;
    for (int i=1;i<=n;++i) data[i].a=x1=read(),data[i].b=x2=read(),mx1=max(mx1,x1),mx2=max(mx2,x2);
    nn=sqrt(mx1);sort(data+1,data+n+1);for (int i=1;i<=mx1;++i) b[i]=(i-1)/nn+1;int now=1;
    for (int i=1;i<=b[mx1];++i) left[i]=(i-1)*nn+1,right[i]=min(mx1,i*nn),ql[i]=1;
    for (int i=0;i<=mx2+1;++i){
        while(now<=n&&data[now].b<i) add(data[now++].a);ll ans=0;int ans1=0;
        for (int j=1;j<=b[mx1];++j){int l=q[j][ql[j]];if (d[l]+(ll)cnt[j]*l>ans) ans=d[l]+(ll)cnt[j]*l,ans1=l;}
        printf("%lld %d\n",ans+(ll)w*i*(n-now+1),ans1);
    }
    return 0;
}