BZOJ 3922 - Karin的弹幕

Karin的弹幕

Problem's Link

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Mean:

给定一个长度为n(1≤n≤70000)序列，有m(1≤m≤70000)次操作：

1. 对一段下标是等差数列的子序列求最大值；

2. 单点修改.

analyse:

如果公差很大，那么速度是很快的。所以我们考虑阈值.

Time complexity: O(N)

view code

/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
*       Author: crazyacking
*       Date  : 2016-02-15-13.19
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long(LL);
typedef unsigned long long(ULL);
const double eps(1e-);

const int N=, MXD=, oo=(~0u>>)+;
int a[N], D, sz[MXD+][MXD+], b[N];
struct node *null;
struct node
{
   node *c[];
   int mx;
   node()
   {
       c[]=c[]=;
       mx=oo;
   }
   void up()
   {
       mx=max(c[]->mx, c[]->mx);
   }
} Po[], *iT=Po, *root[MXD+][MXD+];
node *newnode()
{
   return iT++;
}
node* build(int l, int r)
{
   node *x=newnode();
   if(l==r)
   {
       x->mx=b[l];
       return x;
   }
   int mid=(l+r)>>;
   x->c[]=build(l, mid);
   x->c[]=build(mid+, r);
   x->up();
   return x;
}
int query(int L, int R, int l, int r, node *x)
{
   if(L<=l && r<=R)
   {
       return x->mx;
   }
   int mid=(l+r)>>, ret=oo;
   if(L<=mid)
   {
       ret=query(L, R, l, mid, x->c[]);
   }
   if(mid<R)
   {
       ret=max(ret, query(L, R, mid+, r, x->c[]));
   }
   return ret;
}
void update(int p, int go, int l, int r, node *x)
{
   if(l==r)
   {
       x->mx+=go;
       return;
   }
   int mid=(l+r)>>;
   if(p<=mid)
   {
       update(p, go, l, mid, x->c[]);
   }
   else
   {
       update(p, go, mid+, r, x->c[]);
   }
   x->up();
}
void init(int n)
{
   D=min(MXD, n);
   for(int d=; d<=D; ++d)
   {
       for(int i=; i<=d; ++i)
       {
           int &s=sz[d][i];
           for(int j=i; j<=n; j+=d)
           {
               b[++s]=a[j];
           }
           root[d][i]=build(, s);
       }
   }
}
int query(int x, int d, int n)
{
   if(d>D)
   {
       int mx=oo;
       for(int i=x; i<=n; i+=d)
       {
           mx=max(mx, a[i]);
       }
       return mx;
   }
   int begin=(x-)%d+, pos=(x-)/d+, len=sz[d][begin];
   return query(pos, len, , len, root[d][begin]);
}
void update(int x, int y)
{
   a[x]+=y;
   for(int d=; d<=D; ++d)
   {
       int begin=(x-)%d+, pos=(x-)/d+, len=sz[d][begin];
       update(pos, y, , len, root[d][begin]);
   }
}
int main()
{
   int n, m;
   scanf("%d", &n);
   for(int i=; i<=n; ++i)
   {
       scanf("%d", &a[i]);
   }
   init(n);
   scanf("%d", &m);
   while(m--)
   {
       int op, x, y;
       scanf("%d%d%d", &op, &x, &y);
       if(op)
       {
           printf("%d\n", query(x, y, n));
       }
       else
       {
           update(x, y);
       }
   }
   return ;
}