HDU 3265 Posters

pid=3265" target="_blank" style="">题目链接

题意：给定一些矩形海报。中间有孔。求贴海报的之后的海报覆盖面积并

思路：海报一张能够分割成4个矩形。然后就是普通的矩形面积并了，利用线段树维护就可以

代码：

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

typedef long long ll;

const int N = 50005;

struct Node {

	int l, r, len, cover;

	int size() {return r - l + 1;}

} node[N * 4];

struct Line {

	int l, r, y, flag;

	Line() {}

	Line(int l, int r, int y, int flag) {

		this->l = l; this->r = r;

		this->y = y; this->flag = flag;

	}

} line[N * 8];

struct Rec {

	int x1, y1, x2, y2;

	Rec() {}

	Rec(int x1, int y1, int x2, int y2) {

		this->x1 = x1; this->y1 = y1;

		this->x2 = x2; this->y2 = y2;

	}

} rec[N * 4];

bool cmp(Line a, Line b) {

	return a.y < b.y;

}

int n;

int x[4], y[4];

#define lson(x) ((x<<1)+1)

#define rson(x) ((x<<1)+2)

void pushup(int x) {

	if (node[x].cover) node[x].len = node[x].size();

	else if (node[x].l == node[x].r) node[x].len = 0;

	else node[x].len = node[lson(x)].len + node[rson(x)].len;

}

void build(int l, int r, int x = 0) {

	node[x].l = l; node[x].r = r;

	if (l == r) {

		node[x].cover = node[x].len = 0;

		return;

	}

	int mid = (l + r) / 2;

	build(l, mid, lson(x));

	build(mid + 1, r, rson(x));

	pushup(x);

}

void add(int l, int r, int v, int x = 0) {

	if (l > r) return;

	if (node[x].l >= l && node[x].r <= r) {

		node[x].cover += v;

		pushup(x);

		return;

	}

	int mid = (node[x].l + node[x].r) / 2;

	if (l <= mid) add(l, r, v, lson(x));

	if (r > mid) add(l, r, v, rson(x));

	pushup(x);

}

int main() {

	while (~scanf("%d", &n) && n) {

		build(0, 50000);

		int rn = 0, ln = 0;

		for (int i = 0; i < n; i++) {

			for (int j = 0; j < 4; j++)

				scanf("%d%d", &x[j], &y[j]);

			rec[rn++] = Rec(x[0], y[0], x[1], y[2]);

			rec[rn++] = Rec(x[0], y[2], x[2], y[3]);

			rec[rn++] = Rec(x[0], y[3], x[1], y[1]);

			rec[rn++] = Rec(x[3], y[2], x[1], y[3]);

		}

		for (int i = 0; i < rn; i++) {

			line[ln++] = Line(rec[i].x1, rec[i].x2, rec[i].y1, 1);

			line[ln++] = Line(rec[i].x1, rec[i].x2, rec[i].y2, -1);

		}

		n = ln;

		sort(line, line + n, cmp);

		ll ans = 0;

		for (int i = 0; i < n; i++) {

			if (i) ans += (ll)node[0].len * (line[i].y - line[i - 1].y);

			add(line[i].l, line[i].r - 1, line[i].flag);

		}

		printf("%lld\n", ans);

	}

	return 0;

}