BZOJ 1845: [Cqoi2005] 三角形面积并 [计算几何 扫描线]

1845: [Cqoi2005] 三角形面积并

Time Limit: 3 Sec  Memory Limit: 64 MB
Submit: 1151  Solved: 313
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Description

给出n个三角形，求它们并的面积。

Input

第一行为n(N < = 100)， 即三角形的个数 以下n行，每行6个整数x1, y1, x2, y2, x3, y3，代表三角形的顶点坐标。坐标均为不超过10 ^ 6的实数，输入数据保留1位小数

Output

输出并的面积u, 保留两位小数

Sample Input

2
0.0 0.0 2.0 0.0 1.0 1.0
1.0 0.0 3.0 0.0 2.0 1.0

Sample Output

1.75

---

哈哈哈写出来了 依靠前辈的经验没有卡在-eps上哈哈哈

这是标准的扫描线了吧

找出所有三角形的顶点和交点，排序去重，相邻两个横坐标之间都是梯形或者三角形

梯形面积公式也可以是 中位线*高 

那么就是找这一段区间中位线与三角形交的区间的并了，和圆的面积并一样

一条线与三角形交的区间就不说了很简单

//
//  main.cpp
//  bzoj1845
//
//  Created by Candy on 2017/2/1.
//  Copyright © 2017年 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const double eps=1e-8;
const double INF=1e9;
const int N=1005;

inline int sgn(double x){
    if(abs(x)<eps) return 0;
    else return x<0?-1:1;
}

struct Vector{
    double x,y;
    Vector(double a=0,double b=0):x(a),y(b){}
    bool operator <(const Vector &a)const{
        //return x<a.x||(x==a.x&&y<a.y);
        return sgn(x-a.x)<0||(sgn(x-a.x)==0&&sgn(y-a.y)<0);
    }
    void print(){printf("%lf %lf\n",x,y);}
};
typedef Vector Point;
Vector operator +(Vector a,Vector b){return Vector(a.x+b.x,a.y+b.y);}
Vector operator -(Vector a,Vector b){return Vector(a.x-b.x,a.y-b.y);}
Vector operator *(Vector a,double b){return Vector(a.x*b,a.y*b);}
Vector operator /(Vector a,double b){return Vector(a.x/b,a.y/b);}
bool operator ==(Vector a,Vector b){return sgn(a.x-b.x)==0&&sgn(a.y-b.y)==0;}

double Dot(Vector a,Vector b){return a.x*b.x+a.y*b.y;}
double Cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;}


struct Line{
    Point s,t;
    int id;
    Line(){}
    Line(Point s,Point t,int i):s(s),t(t),id(i){}
    Line(Point s,Point t):s(s),t(t){}
}L[N];
int nl;
Point LI(Line a,Line b){
    Vector v=a.s-b.s,v1=a.t-a.s,v2=b.t-b.s;
    double t=Cross(v2,v)/Cross(v1,v2);
    return a.s+v1*t;
}
bool isLSI(Line l1,Line l2){
    Vector v=l1.t-l1.s,u=l2.s-l1.s,w=l2.t-l1.s;
    return sgn(Cross(v,u))!=sgn(Cross(v,w));
}
struct Triangle{
    Point a,b,c;
    Triangle(){}
    Triangle(Point a,Point b,Point c):a(a),b(b),c(c){}
}t[N];
int n;
double mp[100000];int m;
void iniMP(){
    sort(mp+1,mp+1+m);
    int p=0;
    mp[++p]=mp[1];
    for(int i=2;i<=m;i++) if(mp[i]!=mp[i-1]) mp[++p]=mp[i];
    m=p;
}
Point a,b,c;
struct Interval{
    double l,r;
    bool operator <(const Interval &a) const{
        return l==a.l?r<a.r:l<a.l;
    }
    Interval(double a=0,double b=0):l(a),r(b){}
}in[N];
double solve(double x){
    int m=0;
    for(int i=1;i<=n;i++){
        a=t[i].a,b=t[i].b,c=t[i].c;
        if(sgn(x-min(a.x,min(b.x,c.x)))<0||sgn(x-max(a.x,max(b.x,c.x)))>0 ) continue;
        Line l1(a,b),l2(a,c),l3(b,c),l(Point(x,0),Point(x,1));
        Point p[3];int cnt=0;
        if(isLSI(l,l1)) p[++cnt]=LI(l,l1);
        if(isLSI(l,l2)) p[++cnt]=LI(l,l2);
        if(isLSI(l,l3)&&cnt!=2) p[++cnt]=LI(l,l3);
        in[++m]=Interval(min(p[1].y,p[2].y),max(p[1].y,p[2].y));
    }
    sort(in+1,in+1+m);
    double last=-INF,re=0;
    for(int i=1;i<=m;i++){
        if(in[i].l>last) re+=in[i].r-in[i].l,last=in[i].r;
        else if(in[i].r>last) re+=in[i].r-last,last=in[i].r;
    }
    return re;
}
int main(int argc, const char * argv[]) {
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y);
        t[i]=Triangle(a,b,c);
        mp[++m]=a.x;mp[++m]=b.x;mp[++m]=c.x;
        L[++nl]=Line(a,b,i);L[++nl]=Line(a,c,i);L[++nl]=Line(b,c,i);
    }
    for(int i=1;i<=n*3;i++)
        for(int j=i+1;j<=n*3;j++)
            if(L[i].id!=L[j].id&&sgn(Cross(L[i].t-L[i].s,L[j].t-L[j].s))!=0)
                mp[++m]=LI(L[i],L[j]).x;
    iniMP();
    double ans=0;
    for(int i=1;i<m;i++){
        if(sgn(mp[i+1]-mp[i])>0){
            double x=(mp[i]+mp[i+1])/2;
            ans+=solve(x)*(mp[i+1]-mp[i]);
        }
    }
    printf("%.2lf",ans-eps);
    return 0;
}