Picture POJ - 1177 线段树 扫描线

题解

题目大意 给若干个矩形问覆盖在一起的多边形的边长和是多少
使用扫描线将矩形拆分为线段从上往下扫 用线段树维护线段左右端点是否闭合计算线段分为几段 每次扫描时纵向边长为段数量2高度差 计算横向边时每次减上次的边长减去重复贡献

AC代码

#include <stdio.h>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 10;

struct node2
{
	int l, r, h, f;
	node2() {}
	node2(int a, int b, int c, int d) : l(a), r(b), h(c), f(d) {}
	bool operator < (const node2 &oth) const
	{
		return h < oth.h;
	}
}a[MAXN];
struct node
{
	int l, r, cnt;
	int s, k, lk, rk; //覆盖长度 分几段 左闭合 右闭合
}tre[MAXN * 4];
inline void PushUp(int x)
{
	if (tre[x].cnt)
	{
		tre[x].s = tre[x].r - tre[x].l + 1;
		tre[x].k = tre[x].lk = tre[x].rk = 1;
	}
	else
	{
		tre[x].s = tre[x << 1].s + tre[x << 1 | 1].s;
		tre[x].lk = tre[x << 1].lk, tre[x].rk = tre[x << 1 | 1].rk;
		tre[x].k = tre[x << 1].k + tre[x << 1 | 1].k - (tre[x << 1].rk && tre[x << 1 | 1].lk); //如果两个儿子中心都闭合则答案减1
	}
}
void Build(int x, int l, int r)
{
	tre[x].l = l, tre[x].r = r, tre[x].cnt = tre[x].k = tre[x].lk = tre[x].rk = 0;
	if (l == r)
		return;
	int m = l + r >> 1;
	Build(x << 1, l, m);
	Build(x << 1 | 1, m + 1, r);
}
void Update(int x, int pl, int pr, int v)
{
	int l = tre[x].l, r = tre[x].r;
	if (pl <= l && r <= pr)
	{
		tre[x].cnt += v;
		PushUp(x);
	}
	else
	{
		int m = l + r >> 1;
		if (pl <= m)
			Update(x << 1, pl, pr, v);
		if (pr > m)
			Update(x << 1 | 1, pl, pr, v);
		PushUp(x);
	}
}
int main()
{
#ifdef LOCAL
	freopen("C:/input.txt", "r", stdin);
#endif
	int N;
	cin >> N;
	for (int i = 1; i <= N; i++)
	{
		int x1, y1, x2, y2;
		scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
		x1 += 2e4, y1 += 2e4, x2 += 2e4, y2 += 2e4; //不离散化直接加2e4
		a[i * 2 - 1] = node2(y1, y2, x1, 1);
		a[i * 2] = node2(y1, y2, x2, -1);
	}
	N *= 2;
	ll ans = 0, last = 0;
	sort(a + 1, a + N + 1);
	Build(1, 0, 1e5);
	for (int i = 1; i <= N; i++)
	{
		Update(1, a[i].l, a[i].r - 1, a[i].f);
		if (i != N) //最后一次也要计算
			ans += (a[i + 1].h - a[i].h) * tre[1].k * 2; //计算纵向边长
		ans += abs(tre[1].s - last); //计算横向边长 要减去上次的长度和
		last = tre[1].s;
	}
	cout << ans << endl;

	return 0;
}