HDU 3265 Posters

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：海报一张可以切割成4个矩形，然后就是普通的矩形面积并了，利用线段树维护即可

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;
}