HDU 6638 Snowy Smile 线段树+最大子段和

$n$个点，$-10^9\le x_i,y_i\le 10^9$，求最大子矩阵和。（$n\le 2000$）
Source：2019 Multi-University Training Contest 6

离散化后，枚举矩阵的上下边界，然后将每一列上的数都加入s[y]，则关于该上下边界的最大子矩阵和为$s$数组的最大子段和。
用线段树维护带修改的最大子段和即可，复杂度$O(n^2\log n).$

//std
#include
#include
#include

using namespace std;
typedef long long ll;
const int N = 2010, M = 4100;
int Case, n, m, i, j, k, cb, b[N], pos[N];
ll pre[M], suf[M], s[M], v[M], ans;
struct E {
    int x, y, z;
} e[N];

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

void build(int x, int a, int b) {
    pre[x] = suf[x] = s[x] = v[x] = 0;
    if (a == b) {
        pos[a] = x;
        return;
    }
    int mid = (a + b) >> 1;
    build(x << 1, a, mid), build(x << 1 | 1, mid + 1, b);
}
inline void change(int x, int p) {
    x = pos[x];
    s[x] += p;
    if (s[x] > 0)pre[x] = suf[x] = v[x] = s[x];
    else pre[x] = suf[x] = v[x] = 0;
    for (x >>= 1; x; x >>= 1) {
        pre[x] = max(pre[x << 1], s[x << 1] + pre[x << 1 | 1]);
        suf[x] = max(suf[x << 1 | 1], s[x << 1 | 1] + suf[x << 1]);
        s[x] = s[x << 1] + s[x << 1 | 1];
        v[x] = max(max(v[x << 1], v[x << 1 | 1]), suf[x << 1] + pre[x << 1 | 1]);
    }
}

int main() {
    scanf("%d", &Case);
    while (Case--) {
        scanf("%d", &n);
        for (cb = 0, i = 1; i <= n; i++) {
            scanf("%d%d%d", &e[i].x, &e[i].y, &e[i].z);
            b[++cb] = e[i].y;
        }
        sort(b + 1, b + cb + 1);
        for (m = 0, i = 1; i <= cb; i++)
        	if (i == 1 || b[i] != b[m])b[++m] = b[i];
        sort(e + 1, e + n + 1, cmp);
        ans = 0;
        for (i = 1; i <= n; i++)
        	e[i].y = lower_bound(b + 1, b + m + 1, e[i].y) - b;
        for (i = 1; i <= n; i++)
            if (i == 1 || e[i].x != e[i - 1].x) {
                build(1, 1, m);
                for (j = i; j <= n; j = k) {
                    for (k = j; k <= n && e[j].x == e[k].x; k++)change(e[k].y, e[k].z);
                    if (ans < v[1])ans = v[1];
                }
            }
        printf("%lld\n", ans);
    }
}