三角剖分求多边形面积的交 HDU3060
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三角剖分求多边形面积的交 HDU3060
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1 //三角剖分求多邊形面積的交 HDU3060
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 PolygonArea(a, na) + PolygonArea(b, nb) - res;
95 }
96
97 Point p1[maxn], p2[maxn];
98 int n1, n2;
99
100 int main() {
101 int i;
102 while(scanf("%d%d", &n1, &n2) != EOF) {
103 for(i = 0; i < n1; ++ i) scanf("%lf%lf", &p1[i].x, &p1[i].y);
104 for(i = 0; i < n2; ++ i) scanf("%lf%lf", &p2[i].x, &p2[i].y);
105 printf("%.2f\n", SPIA(p1, p2, n1, n2) + eps);
106 }
107 return 0;
108 }
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轉載于:https://www.cnblogs.com/ITUPC/p/5891030.html
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