HDU 1255 矩形覆盖面积（线段树）

题意：给出若干个矩形，问至少覆盖两次的面积是多少

思路：线段树。。我是直接更新到底，耗时较大，看了网上的200ms能过，我的花了1s，不过代码简洁。。

#include<cstdio>
#include<cstring>
#include<algorithm>
const int maxn = 2 * 1e3 + 10;
using namespace std;

struct mt {
    double l, r, h;
    int add;
    mt() {}
    mt(double l, double r, double h, int a) :
        l(l), r(r), h(h), add(a) {}
    bool operator < (mt p) const { return h < p.h; }
} Mt[2 * maxn];
double x[2 * maxn], seg[4 * maxn];
int cover[4 * maxn];
int n, ql, qr, kase = 1, T;
double X1, X2, Y1, Y2;

void update(int node, int l, int r, int add) {
    if(l == r) {
        cover[node] += add;
        if(cover[node] > 1) seg[node] = x[r + 1] - x[l];
        else seg[node] = 0;
        return ;
    }
    int mid = (l + r) >> 1;
    if(ql <= mid) update(node * 2, l, mid, add);
    if(qr > mid) update(node * 2 + 1, mid + 1, r, add);
    seg[node] = seg[node * 2] + seg[node * 2 + 1];
}

int main() {
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        memset(cover, 0, sizeof(cover));
        memset(seg, 0, sizeof(seg));
        int num = 0;
        for(int i = 0; i < n; i++) {
            scanf("%lf %lf %lf %lf", &X1, &Y1, &X2, &Y2);
            Mt[num] = mt(X1, X2, Y1, 1);
            Mt[num + 1] = mt(X1, X2, Y2, -1);
            x[num] = X1; x[num + 1] = X2;
            num += 2;
        }
        sort(x, x + num);
        sort(Mt, Mt + num);
        int nn = unique(x, x + num) - x;
        double ans = 0;
        for(int i = 0; i < num - 1; i++) {
            ql = lower_bound(x, x + nn, Mt[i].l) - x;
            qr = lower_bound(x, x + nn, Mt[i].r) - x - 1;
            update(1, 0, num - 1, Mt[i].add);
            ans += (Mt[i + 1].h - Mt[i].h) * seg[1];
        }
        printf("%.2f\n", ans);
    }
    return 0;
}