线段树

可以先把纵坐标离散化，然后按横坐标排序，然后按照x来枚举矩形的左右边界，因为排完序之后相同的
x肯定是在一起的。

用线段树维护相同y下的权值，这样问题就变成区间最大子段和。

#include <bits/stdc++.h>
#define INF 2333333333333333333
#define full(a, b) memset(a, b, sizeof a)
#define __fastIn ios::sync_with_stdio(false), cin.tie(0)
#define range(x) (x).begin(), (x).end()
#define pb push_back
using namespace std;
typedef long long LL;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int ret = 0, w = 0; char ch = 0;
    while(!isdigit(ch)){
        w |= ch == '-', ch = getchar();
    }
    while(isdigit(ch)){
        ret = (ret << 3) + (ret << 1) + (ch ^ 48);
        ch = getchar();
    }
    return w ? -ret : ret;
}
template <typename A>
inline A lcm(A a, A b){ return a / __gcd(a, b) * b; }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 3000;
int _, n, x[N], y[N], w[N], a[N];
LL pre[N<<2], suf[N<<2], sum[N<<2], tree[N<<2];
struct Col{
    int x, y, w;
    Col(){}
    Col(int x, int y, int w): x(x), y(y), w(w){}
    bool operator < (const Col &rhs) const {
        return x < rhs.x;
    }
};
vector<Col> v;

void push_up(int rt){
    int l = rt << 1, r = rt << 1 | 1;
    pre[rt] = max(pre[l], sum[l] + pre[r]);
    suf[rt] = max(suf[r], sum[r] + suf[l]);
    sum[rt] = sum[l] + sum[r];
    tree[rt] = max(suf[l] + pre[r], max(tree[l], tree[r]));
}

void buildTree(int rt, int l, int r){
    if(l == r){
        pre[rt] = suf[rt] = sum[rt] = tree[rt] = 0;
        return;
    }
    int mid = (l + r) >> 1;
    buildTree(rt << 1, l, mid);
    buildTree(rt << 1 | 1, mid + 1, r);
    push_up(rt);
}

void insert(int rt, int l, int r, int pos, LL val){
    if(l == r){
        pre[rt] += val, suf[rt] += val, sum[rt] += val, tree[rt] += val;
        return;
    }
    int mid = (l + r) >> 1;
    if(pos <= mid) insert(rt << 1, l, mid, pos, val);
    else insert(rt << 1 | 1, mid + 1, r, pos, val);
    push_up(rt);
}

int main(){

    //freopen("data.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    for(_ = read(); _; _ --){
        n = read();
        for(int i = 1; i <= n; i ++){
            x[i] = read(), y[i] = read(), w[i] = read();
            a[i] = y[i];
        }
        sort(a + 1, a + n + 1);
        int k = (int)(unique(a + 1, a + n + 1) - a - 1);
        for(int i = 1; i <= n; i ++){
            int p = (int)(lower_bound(a + 1, a + k + 1, y[i]) - a);
            v.emplace_back(x[i], p, w[i]);
        }
        sort(range(v));
        LL ans = 0;
        for(int i = 0; i < n; i ++){
            if(i == 0 || v[i].x != v[i - 1].x){
                buildTree(1, 1, k);
                int j = i;
                for(; j < n; j ++){
                    if(v[j].x == v[i].x) insert(1, 1, k, v[j].y, v[j].w);
                    else break;
                }
                ans = max(ans, tree[1]);
                int p;
                for(; j < n; j = p){
                    for(p = j; p < n; p ++){
                        if(v[j].x == v[p].x) insert(1, 1, k, v[p].y, v[p].w);
                        else break;
                    }
                    ans = max(ans, tree[1]);
                }
            }
        }
        printf("%lld\n", ans);
        v.clear();
    }
    return 0;
}

2019 Multi-University Training Contest 6 - Snowy Smile

原文：https://www.cnblogs.com/onionQAQ/p/11319511.html