[USACO5.5]矩形周长Picture[扫描线+线段树]

题意：给出一些矩阵，求这些矩阵合并后外部（被包括在内部的不算）周长

端点-1这个是用点代替了边，区间内有几个点就代表区间长度是多少

#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
inline void chmax(int &x, int y) {if (x < y) x = y;}
inline void chmin(int &x, int y) {if (x > y) x = y;}
int Max = -inf, Min = inf, n, m, tot, ans, lastlen, x, y, xx, yy;
struct Edge {
	int l, r, h, v;
	inline bool operator < (const Edge & rhs) const {
		return h == rhs.h ? v > rhs.v : h < rhs.h;
	}
} G[10005];
inline void add(int l, int r, int h, int v) {
	G[++tot] = (Edge) {l, r, h, v};
}
struct Node {
	int sum, cnt, len;
	//sum区间整体覆盖次数，cnt有几条不相交的线段覆盖了这个区间
	//len区间内被覆盖的长度
	bool rcvd, lcvd;//左右端点是否被覆盖
	Node *ls, *rs;
	inline void pushup(int l, int r) {
		if (sum) cnt = 1, len = r-l+1, lcvd = rcvd = 1;
		else if (l == r) len = 0, cnt = 0, lcvd = rcvd = 0;//Attention
		else {
			len = ls->len + rs->len;
			cnt = ls->cnt + rs->cnt;
			if (ls->rcvd && rs->lcvd) --cnt;
			lcvd = ls->lcvd;
			rcvd = rs->rcvd;
		}
	}
} pool[100005], *root;
void add(Node *cur, int l, int r, int ql, int qr, int v) {
	if (ql <= l && r <= qr) return cur->sum += v, cur->pushup(l, r);
	int mid = l+r>>1;
	if (ql <= mid) add(cur->ls, l, mid, ql, qr, v);
	if (qr >  mid) add(cur->rs, mid+1, r, ql, qr, v);
	cur->pushup(l, r);
}
inline Node *newNode() {
	static int cnt = 0;
	return &(pool[cnt++]);
}
Node *build(int l, int r) {
	Node *cur = newNode();
	if (l == r) return cur;
	int mid = l+r>>1;
	cur->ls = build(l, mid), cur->rs = build(mid+1, r);
	return cur;
}

int main(void) {
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) {
		scanf("%d%d%d%d", &x, &y, &xx, &yy);
		chmax(Max, xx), chmin(Min, x), add(x, xx, y, 1), add(x, xx, yy, -1);
	}
	if (Min <= 0) {//把坐标转为正数，其实直接加上10001(本题坐标最大值)也是可以的
		for(int i = 1; i <= tot; ++i) G[i].l += -Min+1, G[i].r += -Min+1;
		Max -= Min;
	}
	root = build(1, Max);
	sort(G+1, G+1+tot);
	for(int i = 1; i <= tot; ++i) {
		add(root, 1, Max, G[i].l, G[i].r-1, G[i].v);//不是很懂为什么要-1，这奇怪的方式
		while(G[i].h==G[i+1].h&&G[i].v==G[i+1].v) {//防止多次统计答案
			++i;
			add(root, 1, Max, G[i].l, G[i].r-1, G[i].v);
		}
		ans += abs(root->len-lastlen);//统计横边长度
		lastlen = root->len;
		ans += root->cnt*2*(G[i+1].h-G[i].h);//统计竖边长度
	}
	printf("%d\n", ans);
	return 0;
}