题意:

求一个子矩阵要求其矩阵内的合最大。

题解:

正常的求最大子矩阵的复杂度是O(n^3)

对于这一题说复杂度过不去,注意到这个题总共只有2000个点关键点在与这里优化

最大子矩阵可以压缩矩阵变成最大字段和问题

然后可以通过带修改的最大字段和维护这2000个点,复杂度就变成了了O(n^2logn)

将算出每一列的合的操作 用待修改的最大字段和的线段树维护。

 #include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <set>
#include <iostream>
#include <map>
#include <stack>
#include <string>
#include <vector>
#define pi acos(-1.0)
#define eps 1e-9
#define fi first
#define se second
#define rtl rt<<1
#define rtr rt<<1|1
#define bug printf("******\n")
#define mem(a,b) memset(a,b,sizeof(a))
#define name2str(x) #x
#define fuck(x) cout<<#x" = "<<x<<endl
#define f(a) a*a
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define pf printf
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)+
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define FIN freopen("data.txt","r",stdin)
#define gcd(a,b) __gcd(a,b)
#define lowbit(x) x&-x
#define rep(i,a,b) for(int i=a;i<b;++i)
#define per(i,a,b) for(int i=a-1;i>=b;--i)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = + ;
const int maxm = 8e6 + ;
const int mod = 1e9 + ;
const int INF = 0x3f3f3f3f;
int T, n;
LL s[maxn], x[maxn], y[maxn], a[maxn], b[maxn], w[maxn], mp[maxn][maxn];
struct Segtree {
LL maxx, vl, vr, sum, fg; } Tree[maxn << ];
void updata ( int rt ) {
Tree[rt].maxx = max ( Tree[rtl].maxx, max ( Tree[rtr].maxx, Tree[rtl].vr + Tree[rtr].vl ) );
Tree[rt].sum = Tree[rtl].sum + Tree[rtr].sum;
Tree[rt].vl = max ( Tree[rtl].vl, Tree[rtl].sum + Tree[rtr].vl );
Tree[rt].vr = max ( Tree[rtr].vr, Tree[rtr].sum + Tree[rtl].vr );
}
void build ( int l, int r, int rt ) {
Tree[rt].fg = true;
if ( l == r ) {
Tree[rt].sum = s[l];
Tree[rt].maxx = s[l];
Tree[rt].vl = s[l];
Tree[rt].vr = s[l];
return ;
}
int mid = ( l + r ) >> ;
build ( l, mid, rtl );
build ( mid + , r, rtr );
updata ( rt );
}
void add ( int l, int r, int rt, int pos, int to ) {
if ( l > pos || r < pos ) return ;
if ( l == r ) {
Tree[rt].sum += to;
Tree[rt].maxx += to;
Tree[rt].vl += to;
Tree[rt].vr += to;
return ;
}
int mid = ( l + r ) >> ;
add ( l, mid, rtl, pos, to );
add ( mid + , r, rtr, pos, to );
updata ( rt );
}
Segtree query ( int l, int r, int rt, int sa, int se ) {
if ( sa <= l && r <= se ) return Tree[rt];
int mid = ( l + r ) >> ;
if ( sa > mid ) return query ( mid + , r, rtr, sa, se );
if ( se <= mid ) return query ( l, mid, rtl, sa, se );
Segtree t, lson, rson;
lson = query ( l, mid, rtl, sa, se );
rson = query ( mid + , r, rtr, sa, se );
t.vl = max ( lson.vl, lson.sum + rson.vl );
t.vr = max ( rson.vr, lson.vr + rson.sum );
t.maxx = max ( lson.vr + rson.vl, max ( lson.maxx, rson.maxx ) );
return t;
}
vector<LL>v[maxn];
int main() {
// FIN;
sf ( T );
while ( T-- ) {
sf ( n );
for ( int i = ; i <= n ; i++ ) {
scanf ( "%lld%lld%lld", &x[i], &y[i], &w[i] );
a[i] = x[i], b[i] = y[i];
v[i].clear();
}
sort ( a + , a + + n ), sort ( b + , b + + n );
int len1 = unique ( a + , a + + n ) - a - ;
int len2 = unique ( b + , b + + n ) - b - ;
for ( int i = ; i <= n ; i++ ) for ( int j = ; j <= n ; j++ ) mp[i][j] = ;
for ( int i = ; i <= n ; i++ ) {
x[i] = lower_bound ( a + , a + + len1, x[i] ) - a;
y[i] = lower_bound ( b + , b + + len2, y[i] ) - b;
mp[y[i]][x[i]] += w[i];
}
for ( int i = ; i <= n ; i++ )
for ( int j = ; j <= n ; j++ )
if ( mp[i][j] ) v[i].push_back ( j );
LL ans = ;
for ( int i = ; i <= n ; i++ ) {
for ( int j = ; j <= n ; j++ ) s[j] = mp[i][j];
build ( , n, );
ans = max ( ans, query ( , n, , , n ).maxx );
for ( int j = i + ; j <= n ; j++ ) {
for ( int k = ; k < v[j].size() ; k++ ) {
add ( , n, , v[j][k], mp[j][v[j][k]] );
}
ans = max ( ans, query ( , n, , , n ).maxx );
}
}
printf ( "%lld\n", ans );
}
return ;
}

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