扫描线专题 hdu1255

hdu1255

求覆盖至少两次的面积，和直接求覆盖面积比，就是保证cover>1就可以了。

没有进行lazy操作，因为每一次更新伴随着询问，感觉没有必要。982MS水过。

#include <bits/stdc++.h>
#define clr(x,c) memset(x,c,sizeof(x))
using namespace std;
const int N = 20005;

struct ScanLine {
    double x;
    double upY, downY;
    int flag;
    bool operator<(const ScanLine a) const {
        return x < a.x;
    }
    ScanLine() {}
    ScanLine(double x, double y1, double y2, int f) : x(x), upY(y1), downY(y2), flag(f) {}
} line[N];

double tr[N];
int cover[N];
double yy[N];

#define lson (o<<1)
#define rson (o<<1|1)
#define mid ((l+r)>>1)
int yl, yr, v;

double update(int o, int l, int r)
{
    if (yl > r || yr < l) return 0;

    if (l == r) {
        cover[o] += v;
        if (cover[o] > 1) return tr[o] = yy[r + 1] - yy[l];
        return tr[o] = 0;
    }

    if (yl <= mid) update(lson, l, mid);
    if (yr > mid) update(rson, mid + 1, r);

    return tr[o] = tr[lson] + tr[rson];
}

int main()
{
    int n;
    int t;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        int cnt = 0;
        double x1, y1, x2, y2;
        for (int i = 0; i < n; ++i) {
            scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
            line[++cnt] = ScanLine(x1, y2, y1, 1);
            yy[cnt] = y1;
            line[++cnt] = ScanLine(x2, y2, y1, -1);
            yy[cnt] = y2;
        }
        sort(yy + 1, yy + cnt + 1);
        sort(line + 1, line + cnt + 1);
        int len = unique(yy + 1, yy + cnt + 1) - yy - 1;
        clr(cover, 0);
        clr(tr, 0);
        double ans = 0;
        for (int i = 1; i < cnt; ++i) {
            yl = lower_bound(yy+1, yy+len+1, line[i].downY) - yy;
            yr = lower_bound(yy+1, yy+len+1, line[i].upY) - yy - 1;
            v = line[i].flag;
            ans += update(1, 1, len) * (line[i+1].x - line[i].x);
        }
        printf("%.2f\n", ans);
    }
	return 0;
}