Snowy Smile

题目传送门

解题思路

先把y离散化，然后把点按照x的大小进行排序，我们枚举每一种x作为上边界，然后再枚举其对应的每一种下边界。按照这种顺序插入点，这是一个压维的操作，即在线段树中的y位置加上其w，并利用线段树来更新动态的最大子段和。

代码如下

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;

const int N = 2005;

struct T{
    ll x, y, w;
    T(){}
    T(ll x, ll y, ll w): x(x), y(y), w(w){}
}a[N];

bool cmp(const T& a, const T& b)
{
    return a.x < b.x;
}

int b[N];

struct {
    int l, r;
    ll lx, rx, mx;
    ll sum;
}tree[N << 2];

void build(int k, int l, int r)
{
    tree[k].l = l;
    tree[k].r = r;
    tree[k].sum = tree[k].lx = tree[k].rx = tree[k].mx = 0;
    if(l == r)
        return;
    int mid = (l + r) / 2;
    build(2*k, l, mid);
    build(2*k + 1, mid + 1, r);
}

void push_up(int k)
{
    tree[k].sum = tree[2*k].sum + tree[2*k+1].sum;
    tree[k].lx = max(tree[2*k].lx, tree[2*k].sum + tree[2*k+1].lx);
    tree[k].rx = max(tree[2*k+1].rx, tree[2*k+1].sum + tree[2*k].rx);
    tree[k].mx = max(max(tree[2*k].mx, tree[2*k+1].mx), tree[2*k].rx + tree[2*k+1].lx);
}

void insert(int k, int x, int w)
{
    if(tree[k].l == tree[k].r){
        tree[k].sum = tree[k].lx = tree[k].rx = tree[k].mx = tree[k].mx + w;
        return;
    }
    int mid = (tree[k].l + tree[k].r) / 2;
    if(x <= mid)
        insert(2*k, x, w);
    else
        insert(2*k+1, x, w);
    push_up(k);
}

inline ll query()
{
    return tree[1].mx;
}

int main()
{
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while(t --){
        int n;
        cin >> n;
        for(int i = 1; i <= n; i ++){
            ll x, y, w;
            cin >> 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, query());
                insert(1, a[j].y, a[j].w);
            }
            ans = max(ans, query());
        }
        cout << ans << endl;
    }
    return 0;
}

原文地址：https://www.cnblogs.com/whisperlzw/p/11318112.html