zzulioj 2586 矩形面积

之前写过一次，这次是在学习扫描线，顺道想起来这个题目了，那就用扫描线再写一遍吧

扫描线的思想很简单，去网上找几个blog看一下就会了，不多赘述

吐槽一句：评测机毛病真多，我调用个sort明明加了using namespace std; 还提示我需要在sort前面加std，pragma指令也不能用，会报编译错误，自定义排序 boo cmp()的参数写成 p& a就编译错误，必须去掉&符号，之前没那么多毛病啊，难道评测机感冒了？下面这个代码还是上述问题，如果要提交的话请自行删改相应内容。

这个是O()的时间复杂度，可以用线段树优化到O()

#include <bits/stdc++.h>
#pragma warning (disable:4996)
#define mem(a, b) memset(a, b, sizeof a)
#pragma warning (disable:6031)
typedef long long ll;
using namespace std;
const int N = 310;
struct p {
	ll x, y1, y2, mark;
	p() {}
	p(ll x, ll y1, ll y2, ll mark) {this->x = x;this->y1 = y1;this->y2 = y2;this->mark = mark;}
}arr[N];
bool cmp(p& a, p& b) {
	return a.x < b.x;
}
ll dy[N];
ll c[N];// 记录区间出现次数
ll get(ll y) {
	return std::lower_bound(dy + 1, dy + 5, y) - (dy);
}
int main()
{
	ll t;
	scanf("%lld", &t);
	while (t--) {
		ll x1, x2, y1, y2;
		scanf("%lld %lld %lld %lld", &x1, &y2, &x2, &y1);
		arr[1] = p(x1, y1, y2, 1);
		arr[2] = p(x2, y1, y2, -1);
		ll x3, y3, x4, y4;
		scanf("%lld %lld %lld %lld", &x3, &y4, &x4, &y3);
		arr[3] = p(x3, y3, y4, 1);
		arr[4] = p(x4, y3, y4, -1);
		std::sort(arr + 1, arr + 5, cmp);
		dy[1] = y1;
		dy[2] = y2;
		dy[3] = y3;
		dy[4] = y4;
		std::sort(dy + 1, dy + 5);
		ll len = unique(dy + 1, dy + 5) - (dy + 1);// 离散化后的实际长度
		ll last = arr[1].x;
		mem(c, 0);
		ll res = 0;
		for (ll i = get(arr[1].y1); i < get(arr[1].y2); i++) {
			c[i] += arr[1].mark;
		}
		for (ll i = 2; i <= 4; i++) {
			ll now = arr[i].x;
			ll ans = 0;
			for (ll j = 1; j <= len; j++) {
				if (c[j]) {
					ans += dy[j + 1] - dy[j];
				}
			}
			res += (now - last) * ans;
			last = now;
			for (ll j = get(arr[i].y1); j < get(arr[i].y2); j++) {
				c[j] += arr[i].mark;
			}
		}
		printf("%lld\n", res);
	}
	return 0;
}