Poj1151&HDU1542 Atlantis（扫描线+线段树）

题意

给定$n​$个矩形$(x_1,y_1,x_2,y_2)​$，求这$n​$个矩形的面积并

题解

扫描线裸题，可以不用线段树维护，$O(n^2)$是允许的。

#include <cstdio>
#include <cstring>
#include <algorithm>
using std::sort;
using std::unique;
using std::lower_bound;

const int N = 1e2 + 10;
int n, m, tot;
double ans, x1[N], y1[N], x2[N], y2[N], raw[N << 1], tmp[N << 1];
struct Node {
	double x, dy, uy, g; 
	inline bool operator < (const Node &a) const { return x < a.x; }
} cy[N << 1]; int cnt;
double val[N << 3]; int fg[N << 3];

inline void update(int o, int l, int r) {
	if(fg[o]) val[o] = raw[r + 1] - raw[l];
	else if(l == r) val[o] = 0;
	else val[o] = val[o << 1] + val[o << 1 | 1];
}
void modify (int ml, int mr, int k, int o = 1, int l = 1, int r = m) {
	if(l >= ml && r <= mr) {
		fg[o] += k, update(o, l, r);
		return ;
	}
	int mid = (l + r) >> 1, lc = o << 1, rc = lc | 1;
	if(ml <= mid) modify(ml, mr, k, lc, l, mid);
	if(mr > mid) modify(ml, mr, k, rc, mid + 1, r);
	update(o, l, r);
}

int main () {
	while(scanf("%d", &n) != EOF) {
		if(!n) break; ++tot; ans = m = cnt = 0;
		memset(val, 0, sizeof val), memset(fg, 0, sizeof fg);
		for(int i = 1; i <= n; ++i) {
			scanf("%lf%lf%lf%lf", x1 + i, y1 + i, x2 + i, y2 + i);
			tmp[++m] = y1[i], tmp[++m] = y2[i];
		}
		sort(&tmp[1], &tmp[m + 1]); m = unique(&tmp[1], &tmp[m + 1]) - tmp - 1;
		for(int i = 1; i <= n; ++i) {
			int ind1 = lower_bound(&tmp[1], &tmp[m + 1], y1[i]) - tmp;
			int ind2 = lower_bound(&tmp[1], &tmp[m + 1], y2[i]) - tmp;
			raw[ind1] = y1[i], raw[ind2] = y2[i], y1[i] = ind1, y2[i] = ind2;
		}
		for(int i = 1; i <= n; ++i) {
			cy[++cnt] = (Node){x1[i], y1[i], y2[i], 1};
			cy[++cnt] = (Node){x2[i], y1[i], y2[i], -1};
		} sort(&cy[1], &cy[cnt + 1]);
		for(int i = 1; i <= cnt; ++i) {
			modify(cy[i].dy, cy[i].uy - 1, cy[i].g);
			ans += val[1] * (cy[i + 1].x - cy[i].x);
		}
		printf("Test case #%d\nTotal explored area: %.2lf\n\n", tot, ans);
	}
	return 0;
}