POJ 1389 Area of Simple Polygons 扫描线+线段树面积并

Posted 2020-09-29 ( m Lweleth)

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LINK

题意：同POJ1151

思路：

 

/** @Date    : 2017-07-19 13:24:45
  * @FileName: POJ 1389 线段树+扫描线+面积并 同1151.cpp
  * @Platform: Windows
  * @Author  : Lweleth ([email protected])
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <utility>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <stack>
#include <queue>
#include <math.h>
//#include <bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;


struct yuu
{
	int l, r;
	int flag;
	int ll, rr;
	int len;
}tt[N];

struct line{
	int y, x1, x2;
	int flag;
	line(){}
	line(int _x1, int _x2, int _y, int f){
		x1 = _x1;
		x2 = _x2;
		y = _y;
		flag = f;
	}
};
line li[N];
int cmp(line a, line b)
{
	return a.y < b.y;
}
double a[N], b[N];


void pushup(int p)
{
	if(tt[p].flag > 0)
	{
		tt[p].len = tt[p].rr - tt[p].ll;
		return ;
	}
	if(tt[p].l == tt[p].r - 1)
		tt[p].len = 0;
	else 
		tt[p].len = tt[p << 1].len + tt[p << 1 | 1].len;
}
void build(int l, int r, int p)
{
	tt[p].l = l;
	tt[p].r = r;
	tt[p].len = tt[p].flag = 0;
	tt[p].ll = a[l];
	tt[p].rr = a[r];
	if(l == r - 1)
		return ;
	int mid = (l + r) >> 1;
	build(l, mid, p << 1);
	build(mid, r, p << 1 | 1);
}

void update(int x1, int x2, int flag, int p)
{
	if(x1 == tt[p].ll && x2 == tt[p].rr)
	{
		tt[p].flag += flag;
		pushup(p);
		return ;
	}
	if(x2 <= tt[p << 1].rr)
		update(x1, x2, flag, p << 1);
	else if(x1 >= tt[p << 1 | 1].ll)
		update(x1, x2, flag, p << 1 | 1);
	else
	{
		update(x1, tt[p << 1].rr, flag, p << 1);
		update(tt[p << 1 | 1].ll, x2, flag, p << 1 | 1);
	}
	pushup(p);
}

int main()
{
		int x1, x2, y1, y2;
		while(~scanf("%d%d%d%d", &x1, &y1, &x2, &y2) && ( (~x1) || (~x2) || (~y1) || (~y2)))
		{
			int cnt = 1;
			li[cnt] = line(x1, x2, y1, 1);
			a[cnt++] = x1;
			li[cnt] = line(x1, x2, y2, -1);
			a[cnt++] = x2;
			while(scanf("%d%d%d%d", &x1, &y1, &x2, &y2) && (((~x1)||(~x2)||(~y1)||(~y2))))
			{
				li[cnt] = line(x1, x2, y1, 1);
				a[cnt++] = x1;
				li[cnt] = line(x1, x2, y2, -1);
				a[cnt++] = x2;
			}
			sort(li + 1, li + cnt, cmp);
			sort(a + 1, a + cnt);
			build(1, cnt - 1, 1);
			update(li[1].x1, li[1].x2, li[1].flag, 1);
			int ans = 0;
			for(int i = 2; i < cnt; i++)
			{
				//cout << li[i].x1 << "#" << li[i].x2 << "#" << li[i].y << endl;
				ans += tt[1].len * (li[i].y - li[i - 1].y);
				update(li[i].x1, li[i].x2, li[i].flag, 1);
				//cout << "~";
			}
			printf("%d\n", ans);
			
		}
    return 0;
}

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