2019 Hang electric multi-school Snowy Smile hdu6638 (the largest sub-segment and segment tree)

Meaning of the questions: scattered at some point, there are positive and negative value, a rectangular box, so that the sum of the maximum value of the rectangle

Solution: first discrete y, x sorting order after the insertion point, each of the different enumeration x coordinate as the boundary, the lower boundary of the enumeration, updates the maximum segment tree and sub-segment

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<math.h>
#include<vector>
#include<stack>
#include<string>
#include<stdio.h>

using namespace std;
typedef long long LL;
const int MAXN = 2e3 + 10;

const int maxn=5e4 + 10;

struct T{
    LL x,y,w;
    T(){}
    T(LL x,LL y,LL w) : x(x),y(y),w(w){}
}a[MAXN];

struct Segtree{
    int l,r;
    LL lx,rx,mx,sum;
}st[MAXN << 2];

bool cmp(T a,T b)
{
    return a.x < b.x;
}
LL b[MAXN];

void pushup(int o)
{
    st[o].sum = st[o << 1].sum + st[o << 1 | 1].sum;
    st[o].lx = max(st[o << 1].lx,st[o << 1].sum + st[o << 1 | 1].lx);
    st[o].rx = max(st[o << 1 | 1].rx,st[o << 1 | 1].sum + st[o << 1].rx);
    st[o].mx = max(max(st[o << 1].mx,st[o << 1 | 1].mx),st[o << 1].rx + st[o << 1 | 1].lx);
}
void build(int o,int l,int r)
{
    st[o].l = l; st[o].r = r;
    st[o].sum = st[o].mx = st[o].lx = st[o].rx = 0;
    if(l == r)
    {
        return;
    }
    int m = (l + r) >> 1;
    build(o << 1, l, m);
    build(o << 1 | 1, m + 1, r);
}
void insert(int o,int x,int w)
{
    if(st[o].l == st[o].r)
    {
        st[o].sum = st[o].lx = st[o].rx = st[o].mx = st[o].mx + w;
        return;
    }
    int m = (st[o].l + st[o].r) >> 1;
    if(x <= m) insert(o << 1,x,w);
    else insert(o << 1 | 1,x,w);
    pushup(o);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        for(int i = 1; i <= n; i++)
        {
            LL x,y,w;
            scanf("%lld %lld %lld",&x,&y,&w);
            a[i] = T(x,y,w);
            b[i] = y;
        }
        sort(b + 1,b + n + 1);
        int k = unique(b + 1,b + n + 1) - b - 1;
        for(int i = 1; i <= n; i++)
        {
            int y = lower_bound(b + 1,b + k + 1,a[i].y) - b;
            a[i].y = y;
        }
        sort(a + 1,a + n + 1,cmp);
        LL ans = 0;
        for(int i = 1; i <= n; i++)
        {
            if(i != 1 && a[i].x == a[i - 1].x)
                continue;
            build(1,1,k);
            for(int j = i; j <= n; j++)
            {
                if(j != i && a[j].x != a[j - 1].x)
                    ans = max(ans,st[1].mx);
                insert(1,a[j].y,a[j].w);
            }
            ans = max(ans,st[1].mx);
        }
        cout << ans << endl;
    }
}