• (三角剖分)三角形和矩形的面积交 2016湖南省赛


      1 // (三角剖分)三角形和矩形的面积交 2016湖南省赛
      2 
      3 #include <iostream>
      4 #include <cstdio>
      5 #include <cstring>
      6 #include <stack>
      7 #include <queue>
      8 #include <cmath>
      9 #include <algorithm>
     10 using namespace std;
     11 
     12 const int maxn = 555;
     13 const int maxisn = 10;
     14 const double eps = 1e-8;
     15 const double pi = acos(-1.0);
     16 
     17 int dcmp(double x) {
     18     if(x > eps) return 1;
     19     return x < -eps ? -1 : 0;
     20 }
     21 
     22 struct Point {
     23     double x, y;
     24     Point() {
     25         x = y = 0;
     26     }
     27     Point(double a, double b) {
     28         x = a, y = b;
     29     }
     30     inline Point operator-(const Point &b)const {
     31         return Point(x - b.x, y - b.y);
     32     }
     33     inline Point operator+(const Point &b)const {
     34         return Point(x + b.x, y + b.y);
     35     }
     36     inline double dot(const Point &b)const {
     37         return x * b.x + y * b.y;
     38     }
     39     inline double cross(const Point &b, const Point &c)const {
     40         return (b.x - x) * (c.y - y) - (c.x - x) * (b.y - y);
     41     }
     42 };
     43 
     44 Point LineCross(const Point &a, const Point &b, const Point &c, const Point &d) {
     45     double u = a.cross(b, c), v = b.cross(a, d);
     46     return Point((c.x * v + d.x * u) / (u + v), (c.y * v + d.y * u) / (u + v));
     47 }
     48 
     49 double PolygonArea(Point p[], int n) {
     50     if(n < 3) return 0.0;
     51     double s = p[0].y * (p[n - 1].x - p[1].x);
     52     p[n] = p[0];
     53     for(int i = 1; i < n; ++ i)
     54         s += p[i].y * (p[i - 1].x - p[i + 1].x);
     55     return fabs(s * 0.5);
     56 }
     57 
     58 double CPIA(Point a[], Point b[], int na, int nb) { //ConvexPolygonIntersectArea
     59     Point p[maxisn], tmp[maxisn];
     60     int i, j, tn, sflag, eflag;
     61     a[na] = a[0], b[nb] = b[0];
     62     memcpy(p, b, sizeof(Point) * (nb + 1));
     63     for(i = 0; i < na && nb > 2; ++ i) {
     64         sflag = dcmp(a[i].cross(a[i + 1], p[0]));
     65         for(j = tn = 0; j < nb; ++ j, sflag = eflag) {
     66             if(sflag >= 0) tmp[tn ++] = p[j];
     67             eflag = dcmp(a[i].cross(a[i + 1], p[j + 1]));
     68             if((sflag ^ eflag) == -2)
     69                 tmp[tn ++] = LineCross(a[i], a[i + 1], p[j], p[j + 1]);
     70         }
     71         memcpy(p, tmp, sizeof(Point) * tn);
     72         nb = tn, p[nb] = p[0];
     73     }
     74     if(nb < 3) return 0.0;
     75     return PolygonArea(p, nb);
     76 }
     77 
     78 double SPIA(Point a[], Point b[], int na, int nb) { //SimplePolygonIntersectArea
     79     int i, j;
     80     Point t1[4], t2[4];
     81     double res = 0, if_clock_t1, if_clock_t2;
     82     a[na] = t1[0] = a[0], b[nb] = t2[0] = b[0];
     83     for(i = 2; i < na; ++ i) {
     84         t1[1] = a[i - 1], t1[2] = a[i];
     85         if_clock_t1 = dcmp(t1[0].cross(t1[1], t1[2]));
     86         if(if_clock_t1 < 0) std::swap(t1[1], t1[2]);
     87         for(j = 2; j < nb; ++ j) {
     88             t2[1] = b[j - 1], t2[2] = b[j];
     89             if_clock_t2 = dcmp(t2[0].cross(t2[1], t2[2]));
     90             if(if_clock_t2 < 0) std::swap(t2[1], t2[2]);
     91             res += CPIA(t1, t2, 3, 3) * if_clock_t1 * if_clock_t2;
     92         }
     93     }
     94     return fabs(res);
     95 }
     96 
     97 Point p1[maxn], p2[maxn];
     98 int n1, n2;
     99 
    100 int main() {
    101     double x1,x2,x3,x4,y1,y2,y3,y4;
    102     while(~scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2)){
    103         scanf("%lf%lf%lf%lf",&x3,&y3,&x4,&y4);
    104         p1[0].x=x1,p1[0].y=y1;
    105         p1[1].x=x1,p1[1].y=y2;
    106         p1[2].x=x2,p1[2].y=y1;
    107         p2[0].x=x3,p2[0].y=y3;
    108         p2[1].x=x3,p2[1].y=y4;
    109         p2[2].x=x4,p2[2].y=y4;
    110         p2[3].x=x4,p2[3].y=y3;
    111         double ans=SPIA(p1,p2,3,4);
    112         printf("%.6f
    ",ans);
    113     }
    114     return 0;
    115 }
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  • 原文地址:https://www.cnblogs.com/ITUPC/p/5891077.html
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