BZOJ 3922

Karin的弹幕

Problem's Link

 ----------------------------------------------------------------------------

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-8);

const int N=70005, MXD=10, oo=(~0u>>1)+1;
int a[N], D, sz[MXD+1][MXD+1], b[N];
struct node *null;
struct node
{
   node *c[2];
   int mx;
   node()
   {
       c[0]=c[1]=0;
       mx=oo;
   }
   void up()
   {
       mx=max(c[0]->mx, c[1]->mx);
   }
} Po[10000005], *iT=Po, *root[MXD+1][MXD+1];
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)>>1;
   x->c[0]=build(l, mid);
   x->c[1]=build(mid+1, 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)>>1, ret=oo;
   if(L<=mid)
   {
       ret=query(L, R, l, mid, x->c[0]);
   }
   if(mid<R)
   {
       ret=max(ret, query(L, R, mid+1, r, x->c[1]));
   }
   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)>>1;
   if(p<=mid)
   {
       update(p, go, l, mid, x->c[0]);
   }
   else
   {
       update(p, go, mid+1, r, x->c[1]);
   }
   x->up();
}
void init(int n)
{
   D=min(MXD, n);
   for(int d=1; d<=D; ++d)
   {
       for(int i=1; i<=d; ++i)
       {
           int &s=sz[d][i];
           for(int j=i; j<=n; j+=d)
           {
               b[++s]=a[j];
           }
           root[d][i]=build(1, 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-1)%d+1, pos=(x-1)/d+1, len=sz[d][begin];
   return query(pos, len, 1, len, root[d][begin]);
}
void update(int x, int y)
{
   a[x]+=y;
   for(int d=1; d<=D; ++d)
   {
       int begin=(x-1)%d+1, pos=(x-1)/d+1, len=sz[d][begin];
       update(pos, y, 1, len, root[d][begin]);
   }
}
int main()
{
   int n, m;
   scanf("%d", &n);
   for(int i=1; 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
", query(x, y, n));
       }
       else
       {
           update(x, y);
       }
   }
   return 0;
}