CSP-J 2020 T4 Grid Access

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For a DP question, the standard practice is $O\left(n^2\right)$The monotonic queue optimizes DP, but line segment trees can also be used to optimize DP

Carefully observe the way the bear walks and find that he cannot walk to the left. In other words, when he reaches the $When\; j$ , you can never go back to the first$1\dots j-1$ column went. So, we can follow the columns to DP

设 $f_{i,j}$Means to go to $(i,j)$ The maximum value of the sum of integers that can be taken
Transfer:
$f_{i,j}=\underset{k\le i}{\max }{f}_{k,j-1}+s_{i,j}-s_{k-1,j}$
$f_{i,j}=\underset{k>i}{\max }{f}_{k,j-1}+s_{k,j}-s_{i-1,j}$
Finally, use a monotonic queue or line segment tree to optimize it. The
code is written as a line segment tree.

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const long long Maxn=1010;
const long long Maxm=4040;
const long long inf=(1ll<<60);
long long f[Maxn][Maxn],a[Maxn][Maxn];
long long s[Maxn][Maxn];
long long maxv[Maxm],add[Maxm];
long long n,m;
inline long long read()
{
    
    
	long long s=0,w=1;
	char ch=getchar();
	while(ch<'0' || ch>'9'){
    
    if(ch=='-')w=-1;ch=getchar();}
	while(ch>='0' && ch<='9')s=(s<<3)+(s<<1)+(ch^48),ch=getchar();
	return s*w;
}
inline void check(long long &x,long long y)
{
    
    
	x=max(x,y);
}
inline void push_up(long long k)
{
    
    
	maxv[k]=max(maxv[k<<1],maxv[k<<1|1]);
}
inline void upd(long long k,long long v)
{
    
    
	add[k]+=v;
	maxv[k]+=v;
}
inline void push_down(long long k)
{
    
    
	if(!add[k])return;
	upd(k<<1,add[k]);
	upd(k<<1|1,add[k]);
	add[k]=0ll;
}
void build(long long k,long long l,long long r,long long pos)
{
    
    
	add[k]=0ll;
	if(l==r)
	{
    
    
		maxv[k]=f[l][pos];
		return;
	}
	long long mid=(l+r)>>1;
	build(k<<1,l,mid,pos);
	build(k<<1|1,mid+1,r,pos);
	push_up(k);
}
void modify(long long k,long long l,long long r,long long x,long long y,long long v)
{
    
    
	if(x<=l && r<=y)return upd(k,v);
	push_down(k);
	long long mid=(l+r)>>1;
	if(x<=mid)modify(k<<1,l,mid,x,y,v);
	if(mid<y)modify(k<<1|1,mid+1,r,x,y,v);
	push_up(k);
}
long long query(long long k,long long l,long long r,long long x,long long y)
{
    
    
	if(x<=l && r<=y)return maxv[k];
	push_down(k);
	long long mid=(l+r)>>1,ret=-inf;
	if(x<=mid)ret=query(k<<1,l,mid,x,y);
	if(mid<y)check(ret,query(k<<1|1,mid+1,r,x,y));
	return ret;
}
int main()
{
    
    
//	freopen("number.in","r",stdin);
//	freopen("number.out","w",stdout);
//	freopen("in.txt","r",stdin);
//	freopen("out.txt","w",stdout);
	n=read(),m=read();
	for(long long i=1;i<=n;++i)
	for(long long j=1;j<=m;++j)
	{
    
    
		a[i][j]=read();
		f[i][j]=-inf;
	}
	for(long long j=1;j<=m;++j)
	for(long long i=1;i<=n;++i)
	s[i][j]=s[i-1][j]+a[i][j];
	for(long long i=1;i<=n;++i)
	f[i][1]=s[i][1];
	if(n<=300 && m<=300)
	{
    
    
		for(long long j=2;j<=m;++j)
		for(long long i=1;i<=n;++i)
		for(long long k=1;k<=n;++k)
		{
    
    
			long long x=i,y=k;
			if(x>y)swap(x,y);
			check(f[i][j],f[k][j-1]+s[y][j]-s[x-1][j]);
		}
		printf("%lld\n",f[n][m]);
		return 0;
	}
	for(long long j=2;j<=m;++j)
	{
    
    
		build(1,1,n,j-1);
		for(long long i=1;i<=n;++i)
		{
    
    
			modify(1,1,n,1,i,a[i][j]);
			check(f[i][j],query(1,1,n,1,i));
		}
		build(1,1,n,j-1);
		for(long long i=n;i;--i)
		{
    
    
			modify(1,1,n,i,n,a[i][j]);
			check(f[i][j],query(1,1,n,i,n));
		}
	}
	printf("%lld\n",f[n][m]);
	return 0;
}