HDU 1255 矩形覆盖面积（线段树）

来源：互联网发布：myeclipse2016破解mac 编辑：程序博客网时间：2024/06/06 16:34

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;}

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