【HDU2019多校】E - Snowy Smile (最大字段和）

There are nn pirate chests buried in Byteland, labeled by 1,2,…,n1,2,…,n. The ii-th chest's location is (xi,yi)(xi,yi), and its value is wiwi, wiwi can be negative since the pirate can add some poisonous gases into the chest. When you open the ii-th pirate chest, you will get wiwi value. 

You want to make money from these pirate chests. You can select a rectangle, the sides of which are all paralleled to the axes, and then all the chests inside it or on its border will be opened. Note that you must open all the chests within that range regardless of their values are positive or negative. But you can choose a rectangle with nothing in it to get a zero sum. 

Please write a program to find the best rectangle with maximum total value.

InputThe first line of the input contains an integer T(1≤T≤100)T(1≤T≤100), denoting the number of test cases. 

In each test case, there is one integer n(1≤n≤2000)n(1≤n≤2000) in the first line, denoting the number of pirate chests. 

For the next nn lines, each line contains three integers xi,yi,wi(−109≤xi,yi,wi≤109)xi,yi,wi(−109≤xi,yi,wi≤109), denoting each pirate chest. 

It is guaranteed that ∑n≤10000∑n≤10000.OutputFor each test case, print a single line containing an integer, denoting the maximum total value.Sample Input

2 4 1 1 50 2 1 50 1 2 50 2 2 -500 2 -1 1 5 -1 1 1

Sample Output

100 6

SOLUTION:
都是都是套路，每次枚举矩形的上边界
然后每个下边界加入线段树后，查询最大字段和。

CODE:
#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);elseinsert(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; }

　　

转载于:https://www.cnblogs.com/zhangbuang/p/11386932.html

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