2019 杭电多校第6场 HDU - 6638 Snowy Smile 线段树 最大子段和

题目链接：https://vjudge.net/problem/HDU-6638

题意：n个点，画一个矩形，使得矩形内的所有点的权值最大，求最大权值

题解：首先将坐标离散化到 O(n) 的范围内，方便后续的处理。 将所有点按照y坐标排序，枚举矩形的上边界，然后往后依次加入每行的点，这样就确定了 矩形的上下边界。线段树维护下每个点的权值，则答案即为最大连续子段和，。 时间复杂度 O(n 2 log n)。维护的时候还要记录左连续，右连续，区间和，区间连续最大值，方便在pushup的时候使用。

#include
using namespace std;
typedef long long ll;
const int N = 2010;
struct Point {
    int x, y, w;
    bool operator < (const Point &b) const {
        if(y != b.y) return y < b.y;
        else return x < b.x;
    }
}p[N];
int n, b[N], len1, len2;
struct node {
    int l, r;
    ll maxx, lm, rm, sum;
}tree[N << 2];
void build(int l, int r, int cur) {
    tree[cur].l = l;
    tree[cur].r = r;
    tree[cur].maxx = tree[cur].lm = tree[cur].rm = tree[cur].sum = 0;
    if(l == r) return;
    int mid = (r + l) >> 1;
    build(l, mid, cur << 1);
    build(mid + 1, r, cur << 1 | 1);
}
void pushup(int cur) {
    tree[cur].maxx = max(0LL, max(tree[cur << 1].maxx, tree[cur << 1 | 1].maxx));
    tree[cur].maxx = max(tree[cur].maxx, max(0LL, tree[cur << 1].rm) + max(0LL, tree[cur << 1 | 1].lm));
    
    tree[cur].lm = max(0LL, max(tree[cur << 1].lm, tree[cur << 1].sum + tree[cur << 1 | 1].lm));
    tree[cur].rm = max(0LL, max(tree[cur << 1 | 1].rm, tree[cur << 1 | 1].sum + tree[cur << 1].rm));
    
    tree[cur].sum = tree[cur << 1].sum + tree[cur << 1 | 1].sum;
}
void update(int pos, int cur, int val) {
    if(tree[cur].l == tree[cur].r) {
        tree[cur].maxx += val;
        tree[cur].lm += val;
        tree[cur].rm += val;
        tree[cur].sum += val;
        return;
    }
    if(pos <= tree[cur << 1].r) update(pos, cur << 1, val);
    else update(pos, cur << 1 | 1, val);
    pushup(cur);
}
int main() {
    int T, pos;
    ll ans;
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        len1 = len2 = 0;
        for(int i = 1; i <= n; i++) scanf("%d %d %d", &p[i].x, &p[i].y, &p[i].w), b[++len1] = p[i].x;
        sort(b + 1, b + 1 + len1);
        len1 = unique(b + 1, b + 1 + len1) - (b + 1);
        for(int i = 1; i <= n; i++) p[i].x = lower_bound(b + 1, b + 1 + len1, p[i].x) - b;

        for(int i = 1; i <= n; i++)    b[++len2] = p[i].y;
        sort(b + 1, b + 1 + len2);
        len2 = unique(b + 1, b + 1 + len2) - (b + 1);
        for(int i = 1; i <= n; i++) p[i].y = lower_bound(b + 1, b + 1 + len2, p[i].y) - b;
        sort(p + 1, p + 1 + n);    
        ans = 0;
        for(int i = 1; i <= len2; i++) {
            for(int j = n; j >= 1; j--) {
                if(p[j].y <= i) {
                    pos = j;
                    break;
                }
            }
            build(1, len1, 1);
            for(int j = i; j >= 1; j--) {
                while(pos && p[pos].y >= j) {
                    update(p[pos].x, 1, p[pos].w);
                    pos--;
                }
                ans = max(ans, tree[1].maxx);
            } 
            
        }    
        printf("%lld\n", ans);
    }
    return 0;
}