hdu 2528 Area

2014-07-30

http://acm.hdu.edu.cn/showproblem.php?pid=2528
解题思路：
　　求多边形被一条直线分成两部分的面积分别是多少。因为题目给的直线一定能把多边形分成两部分，所以就不用考虑多边形是否与直线相交。直接求问题。
　　
　　将多边形的每一条边与直线判断是否相交。若相交，就从这点开始计算面积，直到判断到下一个边与直线相交的点。这之间的面积求出来为area2。
　　area1为多边形的总面积。多边形被直线分成的另外一部分面积 = area1 - area2

　　有一个特殊的情况：当直线与多边形的顶点相交时，应该考虑下如何处理

  1 #include<cmath>
 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 using namespace std;

 #define MAXN 30
 #define EPS 0.00000001

 int dcmp(double x){
     if(fabs(x) < EPS)
         return ;
     return x <  ? - : ;
 }

 struct Point{
     double x, y;
     Point(double x = , double y = ): x(x), y(y) {}

     bool operator != (const Point & other){
         return dcmp(x - other.x) !=  || (dcmp(x - other.x) ==  && dcmp(y - other.y) != );
     }
 };

 struct Line{
     Point A, B;
     //Line(Point A = Point(0, 0), Point B = Point(0, 0)): A(A), B(B){}
 };

 typedef Point Vector;

 Vector operator + (Vector A, Vector B){
     return Vector(A.x + B.x, A.y + B.y);
 }

 Vector operator - (Point A, Point B){
     return Vector(A.x - B.x, A.y - B.y);
 }

 Vector operator * (Vector A, double d){
     return Vector(A.x * d, A.y * d);
 }

 Vector operator / (Vector A, double d){
     return Vector(A.x / d, A.y / d);
 }

 double dot(Vector A, Vector B){//点乘
     return A.x * B.x + A.y * B.y;
 }

 double cross(Vector A, Vector B){//叉乘
     return A.x * B.y - A.y * B.x;
 }

 int n;
 Point p[MAXN];
 Line line;

 bool input(){
     scanf("%d", &n );
     if(!n){
         return false;
     }
     for(int i = ; i < n; i++ ){
         scanf("%lf%lf", &p[i].x, &p[i].y );
     }
     scanf("%lf%lf%lf%lf", &line.A.x, &line.A.y, &line.B.x, &line.B.y );
     return true;
 }

 double polygon_area(){//求多边形面积
     double area = ;
     for(int i = ; i < n - ; i++ ){
         area += cross(p[i] - p[], p[i + ] - p[]);
     }
     return fabs(area) * 0.5;
 }

 bool line_segment_intersect(Line L, Point A, Point B, Point &P){//直线和线段相交
     Vector a = A - L.B, b = L.A - L.B, c = B - L.B;
     if(dcmp(cross(a, b)) * dcmp(cross(b, c)) >= ){//若直线和线段相交 求出交点 《算法入门经典训练之南》上的公式
         Vector u = L.A - A;
         double t = cross(A - B, u) / cross(b, A - B);
         P = L.A + b * t;
         return true;
     }
     return false;
 }

 void solve(){
     int flag = ;
     double area1 = polygon_area(), area2 = ;//area1算出多边形总面积
     Point P, T;

     p[n] = p[];
     for(int i = ; i < n; i++ ){
         if(flag ==  && line_segment_intersect(line, p[i], p[i + ], P)){//第一次相交点
             area2 += cross(P, p[i + ]);
             flag++;
         }else if(flag ==  && line_segment_intersect(line, p[i], p[i + ], T) && P != T){//第二次相交点
             area2 += cross(p[i], T);
             area2 += cross(T, P);
             flag++;
             break;
         }else if(flag == ){
             area2 += cross(p[i], p[i + ]);
         }
     }
     area2 = fabs(area2) * 0.5;
     area1 -= area2;//area1 获取多边形另外一半的面积
     if(area1 < area2){//规定大面积在前面 输出
         swap(area1, area2);
     }
     printf("%.0lf %.0lf\n", area1, area2);
 }

 int main(){
     //freopen("data.in", "r", stdin );
     while(input()){
         solve();
     }
     return ;
 }