P5490 【模板】扫描线 $\&\&$ 扫描线

P5490 【模板】扫描线

题目描述
求 n 个矩形的面积并。
输出格式
一行一个正整数，表示 n 个矩形的并集覆盖的总面积。

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エラー発生：线段树开小了， 因为n变成了两倍，线段树就得开4*2=8倍

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扫描线

对每一根扫描线， 维护所截得的长度， 每次乘以两根扫描线高度差就得到了面积并
截得长度用线段树维护即可
注意线段树需要离散化

Code

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<climits>
#define LL long long
#define REP(i, x, y) for(LL i = (x);i <= (y);i++)
using namespace std;
LL RD(){
    LL out = 0,flag = 1;char c = getchar();
    while(c < '0' || c >'9'){if(c == '-')flag = -1;c = getchar();}
    while(c >= '0' && c <= '9'){out = out * 10 + c - '0';c = getchar();}
    return flag * out;
    }
const LL maxn = 100010;
LL num, X[maxn << 1];
LL tot;//去重后的横坐标数
struct ScanLine{
	LL h, l, r, mark;
	bool operator < (const ScanLine &x) const {
		return h < x.h;
		}
	}line[maxn << 1];
#define lid (id << 1)
#define rid (id << 1) | 1
struct Seg_Tree{
	LL l, r, tag, len;
	}tree[maxn << 4];
void pushup(LL id){
	LL l = tree[id].l, r = tree[id].r;
	if(tree[id].tag)
		tree[id].len = X[r + 1] - X[l];
	else
		tree[id].len = tree[lid].len + tree[rid].len;
	}
void build(LL id, LL l, LL r){
	tree[id].l = l, tree[id].r = r;
	if(l == r){
		tree[id].tag = tree[id].len = 0;
		return ;
		}
	LL mid = (l + r) >> 1;
	build(lid, l, mid), build(rid, mid + 1, r);
	pushup(id);
	}
void update(LL id, LL L, LL R, LL val){
	LL l = tree[id].l, r = tree[id].r;
	if(X[l] >= R || X[r + 1] <= L)return ;
	if(X[l] >= L && X[r + 1] <= R){
		tree[id].tag += val;
		pushup(id);
		return ;
		}
	update(lid, L, R, val);
	update(rid, L, R, val);
	pushup(id);
	}
void init(){
	num = RD();
	REP(i, 1, num){
		LL x1 = RD(), y1 = RD(), x2 = RD(), y2 = RD();
		X[i * 2 - 1] = x1, X[i << 1] = x2;
		line[i * 2 - 1] = (ScanLine){y1, x1, x2, 1};
		line[i << 1] = (ScanLine){y2, x1, x2, -1};
		}
	num = num << 1;
	sort(X + 1, X + 1 + num);
	sort(line + 1, line + 1 + num);
	tot = unique(X + 1, X + 1 + num) - X - 1;
	build(1, 1, tot - 1);
	
	}
void work(){
	LL ans = 0;
	REP(i, 1, num - 1){
		update(1, line[i].l, line[i].r, line[i].mark);
		ans += tree[1].len * (line[i + 1].h - line[i].h);
		}
	cout<<ans<<endl;
	}
int main(){
	init();
	work();
	return 0;
	}