bzoj3594[Scoi2014]方伯伯的玉米田

http://www.lydsy.com/JudgeOnline/problem.php?id=3594

题目就是问你至多操作K次后的最长上升子序列。

首先，我们会得到一个结论：每次操作区间的右端点一定是n。

记$F[i][j]$表示前$i$棵玉米，使用$j$次操作时，以第$i$棵玉米为结尾的最长上升子序列的长度,则：

$$F[i][j]=Max\{F[k][l]|k<i,l\leq j,a[k]+l \leq a[i]+j\}+1$$

状态$F[i][j]$等价于一个三元组$(i,a[i]+j,j)$

其实就是求三元组的三维偏序。

然后我脑抽用CDQ分治，结果TLE。。。。。。~OTATO~

发现其实用二维树状数组即可。

真是*了狗了。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<utility>
#include<set>
#include<bitset>
#include<vector>
#include<functional>
#include<deque>
#include<cctype>
#include<climits>
#include<complex>
//#include<bits/stdc++.h>适用于CF,UOJ，但不适用于poj
 
using namespace std;

typedef long long LL;
typedef double DB;
typedef pair<int,int> PII;
typedef complex<DB> CP;

#define mmst(a,v) memset(a,v,sizeof(a))
#define mmcy(a,b) memcpy(a,b,sizeof(a))
#define fill(a,l,r,v) fill(a+l,a+r+1,v)
#define re(i,a,b)  for(i=(a);i<=(b);i++)
#define red(i,a,b) for(i=(a);i>=(b);i--)
#define ire(i,x) for(typedef(x.begin()) i=x.begin();i!=x.end();i++)
#define fi first
#define se second
#define m_p(a,b) make_pair(a,b)
#define p_b(a) push_back(a)
#define SF scanf
#define PF printf
#define two(k) (1<<(k))

template<class T>inline T sqr(T x){return x*x;}
template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;}
template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;}

inline int sgn(DB x){if(abs(x)<1e-9)return 0;return(x>0)?1:-1;}
const DB Pi=acos(-1.0);

int gint()
  {
        int res=0;bool neg=0;char z;
        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
        if(z==EOF)return 0;
        if(z=='-'){neg=1;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
        return (neg)?-res:res; 
    }
LL gll()
  {
      LL res=0;bool neg=0;char z;
        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
        if(z==EOF)return 0;
        if(z=='-'){neg=1;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
        return (neg)?-res:res; 
    }

const int maxn=10000;
const int maxK=500;
const int maxv=5000+maxK;

int n,K;
int a[maxn+100],v;
int f[maxn+100][maxK+10],ans;

#define lowbit(a) ((a)&(-a))
int tr[maxv+100][maxK+100];
void update(int x,int y,int val)
  {
      int i,j;
      for(i=x;i<=v;i+=lowbit(i))
        for(j=y+1;j<=K+1;j+=lowbit(j))
          upmax(tr[i][j],val);
  }
int ask(int x,int y)
  {
      int res=0,i,j;
      for(i=x;i>=1;i-=lowbit(i))
        for(j=y+1;j>=1;j-=lowbit(j))
          upmax(res,tr[i][j]);
      return res;
  }

int main()
  {
      freopen("bzoj3594.in","r",stdin);
      freopen("bzoj3594.out","w",stdout);
      int i,j;
      n=gint();K=gint();
      re(i,1,n)a[i]=gint(),upmax(v,a[i]);
      v+=K;
      re(i,1,n)
        {
            re(j,0,K)f[i][j]=ask(a[i]+j,j)+1;
            re(j,0,K)update(a[i]+j,j,f[i][j]);
            re(j,0,K)upmax(ans,f[i][j]);
        }
      cout<<ans<<endl;
      return 0;
  }