比赛的时候只在拼命的想怎么去优化O(N^3)的那个之前所认为的标准解法，没想到，这就是一道O(N^2 * logN)的扫描线。
我们可以固定上下两个区间，然后在固定的区域中，就是一维的空间了，我们直接在这一维里去查询即可，然后去寻找区间最大连续子段即可，很方便的。
区间最大连续子段：
我们要查最大左端点，最大右端点，还有整个区间段的和，以及该区间段内的最大值。
然后维护左右段最大连续，都是去比较，子区间的端点最大连续 和 子区间整个和加上另一个子区间的该方向最大连续。
然后区间最大值，就是去比较左右最大值，左右连续，左子区间的右连续和右子区间的左连续之和，这几个的最大值。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MAX_3(x, y, z) max(x, max(y, z))
#define MAX_6(a, b, c, d, f, g) max(max(a, b), max(max(c, d), max(f, g)))
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 2007;
int N, lsan_x[maxN], lsan_y[maxN], len_x, len_y;
vector<int> vt[maxN];
struct node
{
int x, y; ll val;
node(int a=0, int b=0, ll c=0):x(a), y(b), val(c) {}
}a[maxN];
bool cmp(node e1, node e2) { return e1.x == e2.x ? e1.y < e2.y : e1.x < e2.x; }
struct make_Tree
{
ll mx, lm, rm, all;
make_Tree(ll a=0, ll b=0, ll c=0, ll d=0):mx(a), lm(b), rm(c), all(d) {}
}tree[maxN<<2];
void buildTree(int rt, int l, int r)
{
tree[rt] = make_Tree();
if(l == r) return;
int mid = HalF;
buildTree(Lson); buildTree(Rson);
}
inline void pushup(int rt)
{
if(tree[lsn].lm > tree[lsn].all + tree[rsn].lm) tree[rt].lm = tree[lsn].lm;
else tree[rt].lm = tree[lsn].all + tree[rsn].lm;
if(tree[rsn].rm > tree[rsn].all + tree[lsn].rm) tree[rt].rm = tree[rsn].rm;
else tree[rt].rm = tree[rsn].all + tree[lsn].rm;
tree[rt].all = tree[lsn].all + tree[rsn].all;
tree[rt].mx = MAX_6(tree[rt].lm, tree[rt].rm, tree[lsn].mx, tree[rsn].mx, tree[lsn].rm + tree[rsn].lm, tree[rt].all);
}
void update(int rt, int l, int r, int qx, ll val)
{
if(l == r)
{
if(val >= 0) tree[rt] = make_Tree(val, val, val, val);
else tree[rt] = make_Tree(0, 0, 0, val);
return;
}
int mid = HalF;
if(qx <= mid) update(Lson, qx, val);
else update(Rson, qx, val);
pushup(rt);
}
inline ll query(int rt, int l, int r, int ql, int qr)
{
if(ql <= l && qr >= r) return tree[rt].mx;
int mid = HalF;
if(qr <= mid) return query(QL);
else if(ql > mid) return query(QR);
else
{
ll ans = tree[lsn].rm + tree[rsn].lm;
ans = MAX_3(ans, query(QL), query(QR));
return ans;
}
}
int main()
{
int Cas; scanf("%d", &Cas);
while(Cas--)
{
scanf("%d", &N);
for(int i=1; i<=N; i++) vt[i].clear();
for(int i=1; i<=N; i++)
{
scanf("%d%d%lld", &a[i].x, &a[i].y, &a[i].val);
lsan_x[i] = a[i].x; lsan_y[i] = a[i].y;
}
sort(lsan_x + 1, lsan_x + N + 1);
sort(lsan_y + 1, lsan_y + N + 1);
len_x = (int)(unique(lsan_x + 1, lsan_x + N + 1) - lsan_x - 1);
len_y = (int)(unique(lsan_y + 1, lsan_y + N + 1) - lsan_y - 1);
for(int i=1; i<=N; i++)
{
a[i].x = (int)(lower_bound(lsan_x + 1, lsan_x + len_x + 1, a[i].x) - lsan_x);
a[i].y = (int)(lower_bound(lsan_y + 1, lsan_y + len_y + 1, a[i].y) - lsan_y);
}
sort(a + 1, a + N + 1, cmp);
int tot = 0;
for(int i=1, j=0; i<=N; )
{
j = i + 1;
while(a[j].x == a[i].x && a[j].y == a[i].y)
{
a[i].val += a[j].val;
j++;
}
a[++tot] = node(a[i].x, a[i].y, a[i].val);
i = j;
}
for(int i=1; i<=tot; i++)
{
vt[a[i].x].push_back(i);
}
ll ans = 0;
for(int i=1; i<=len_x; i++)
{
buildTree(1, 1, len_y);
ll sum[maxN] = {0};
for(int j=i; j<=len_x; j++)
{
int len = (int)vt[j].size();
for(int k=0, pos; k<len; k++)
{
pos = vt[j][k];
sum[a[pos].y] += a[pos].val;
update(1, 1, len_y, a[pos].y, sum[a[pos].y]);
}
ans = max(ans, query(1, 1, len_y, 1, len_y));
}
}
printf("%lld\n", ans);
}
return 0;
}