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-8using 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;}