标签:include cut oid can 模板 color str operator ios
题意:求多边形的核的面积
套模板即可
#include <iostream> #include <cstdio> #include <cmath> #define eps 1e-18 using namespace std; const int MAXN = 1555; double a, b, c; int n, cnt; struct Point { double x, y; double operator ^(const Point &b) const { return x*b.y - y*b.x; } }point[MAXN], p[MAXN], tp[MAXN]; void Get_equation(Point p1, Point p2) { a = p2.y - p1.y; b = p1.x - p2.x; c = p2.x * p1.y - p1.x * p2.y; } Point Intersection(Point p1, Point p2) { double u = fabs(a * p1.x + b * p1.y + c); double v = fabs(a * p2.x + b * p2.y + c); Point t; t.x = (p1.x * v + p2.x * u) / (u + v); t.y = (p1.y * v + p2.y * u) / (u + v); return t; } void Cut() { int tmp = 0; for(int i=1; i<=cnt; i++) { //顺时针是>-eps和>eps,逆时针是<eps和<-eps if(a * p[i].x + b * p[i].y + c > -eps) tp[++tmp] = p[i]; else { if(a * p[i-1].x + b * p[i-1].y + c > eps) tp[++tmp] = Intersection(p[i-1], p[i]); if(a * p[i+1].x + b * p[i+1].y + c > eps) tp[++tmp] = Intersection(p[i], p[i+1]); } } for(int i=1; i<=tmp; i++) p[i] = tp[i]; p[0] = p[tmp]; p[tmp+1] = p[1]; cnt = tmp; } double solve() { for(int i=1; i<=n; i++) p[i] = point[i]; point[n+1] = point[1]; p[0] = p[n]; p[n+1] = p[1]; cnt = n; for(int i=1; i<=n; i++) { Get_equation(point[i], point[i+1]); Cut(); } double res = 0; for(int i = 1; i <= cnt; i++) res += p[i]^p[i+1]; return fabs(res/2); } int main() { int T; scanf("%d\n", &T); while(T--) { scanf("%d", &n); for(int i=1; i<=n; i++) scanf("%lf%lf", &point[i].x, &point[i].y); printf("%.2f\n", solve()); } return 0; }
POJ 1279 Art Gallery 半平面交 多边形的核
标签:include cut oid can 模板 color str operator ios
原文地址:http://www.cnblogs.com/pach/p/7482240.html