JDFZOJ 1005 多边形面积 扫描线

题目大意：给出N个凸多边形，求这些多边形的面积并。

思路：N只有不到10，乱搞就可以。还有一种更优的解法，似乎只需要O(n^2logn)的时间就可以解决。但是我并不会，想了解的参照：http://wyfcyx.is-programmer.com/posts/80378.html

下面说乱搞的思路。由于都是凸多边形，那么任意一条垂直于x轴的直线在多边形内的区域一定是一条线段（或者什么都没有），那么我们将所有多边形按照梯形进行剖分，求出每个部分的梯形中腰长度并就可以算出总的面积了。

题目网址见：http://oj.jdfz.com.cn:8081/oldoj/problem.php?id=1005

CODE:

#define _CRT_SECURE_NO_WARNINGS

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define MAX 20
#define EPS 1e-8
using namespace std;
#define max(a,b) ((a) > (b) ? (a):(b))
#define min(a,b) ((a) < (b) ? (a):(b))
#define INRANGE(x,y,c) ((c <= y && c >= x) || (c <= x && c >= y))

struct Point{
    double x,y;

    Point(double _,double __):x(_),y(__) {}
    Point() {}
    Point operator +(const Point &a)const {
        return Point(x + a.x,y + a.y);
    }
    Point operator -(const Point &a)const {
        return Point(x - a.x,y - a.y);
    }
    Point operator *(double a)const {
        return Point(x * a,y * a);
    }
    void Read() {
        scanf("%lf%lf",&x,&y);
    }
}temp[1010];

inline double Calc(const Point &p1,const Point &p2)
{
    return sqrt((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y));
}

inline double Cross(const Point &p1,const Point &p2)
{
    return p1.x * p2.y - p1.y * p2.x;
}

struct Segment{
    Point p1,p2,v;

    Segment(Point _,Point __,Point ___):p1(_),p2(__),v(___) {}
    Segment() {}
    bool OnSegment(const Point &p) {
        if(p1.x == p2.x)    return INRANGE(p1.y,p2.y,p.y);
        return INRANGE(p1.x,p2.x,p.x);
    }
}*src[MAX][1010],save[MAX * 1010];
int cnt_seg;

struct Interval{
    double x,y;

    Interval(double _,double __):x(_),y(__) {}
    Interval() {}
    bool operator <(const Interval &a)const {
        if(x == a.x)    return y < a.y;
        return x < a.x;
    }
}interval[MAX * 1010];

inline Point GetIntersection(const Segment &l1,const Segment &l2)
{
    Point u = l1.p1 - l2.p1;
    double t = Cross(l2.v,u) / Cross(l1.v,l2.v);
    return l1.p1 + l1.v * t;
}

inline Segment *MakeSegment(const Point &p1,const Point &p2)
{
    save[++cnt_seg] = Segment(p1,p2,p2 - p1);
    return &save[cnt_seg];
}

inline Interval GetInterval(Segment *src[],double x)
{
    Interval re(0,0);
    for(int i = 1; src[i] != NULL; ++i) {
        Point intersection = GetIntersection(*src[i],Segment(Point(x,0),Point(x,0),Point(0,1)));
        if(src[i]->OnSegment(intersection)) {
            if(!re.x)   re.x = intersection.y;
            else        re.y = intersection.y;
        }
    }
    if(re.x > re.y)  swap(re.x,re.y);
    return re;
}

int cnt;
double divide[1010 * 1010];

int main()
{
    cin >> cnt;
    for(int points,i = 1; i <= cnt; ++i) {
        scanf("%d",&points);
        for(int j = 1; j <= points; ++j)
            temp[j].Read();
        for(int j = 1; j < points; ++j)
            src[i][j] = MakeSegment(temp[j],temp[j + 1]);
        src[i][points] = MakeSegment(temp[points],temp[1]);
    }
    int divides = 0;
    for(int i = 1; i <= cnt_seg; ++i)
        for(int j = i + 1; j <= cnt_seg; ++j) {
            if(fabs(Cross(save[i].v,save[j].v)) < EPS)   continue;
            Point intersection = GetIntersection(save[i],save[j]);
            divide[++divides] = intersection.x;
        }
    sort(divide + 1,divide + divides + 1);
    double area = .0;
    for(int i = 1; i < divides; ++i) {
        double x = (divide[i + 1] + divide[i]) / 2;
        int intervals = 0;
        for(int j = 1; j <= cnt; ++j)
            interval[++intervals] = GetInterval(src[j],x);
        sort(interval + 1,interval + intervals + 1);

        double now = .0,l = interval[1].x,r = interval[1].y;
        for(int j = 2; j <= intervals; ++j)
            if(interval[j].x <= r)
                r = max(r,interval[j].y);
            else {
                now += r - l;
                l = interval[j].x;
                r = interval[j].y;
            }
        now += r - l;
        area += now * (divide[i + 1] - divide[i]);
    }
    cout << fixed << setprecision(3) << area << endl;
    return 0;
}