http://hi.baidu.com/billdu/item/703ad4e15d819db52f140b0b 根据这个思路写的。
将圆与三角形的交按照0、1、2个交点分类
然后就好做多啦，有效面积是个强大的工具
买一送2，poj 3675,2986一样可以用模板过掉。
咔咔。有自己的模板真好~~
#include #include #include #include #include #include #include #include #include #include #include #include #define MID(x,y) ( ( x + y ) >> 1 ) #define L(x) ( x << 1 ) #define R(x) ( x << 1 | 1 ) #define FOR(i,s,t) for(int i=(s); i<(t); i++) #define FORD(i,s,t) for(int i=(s-1); i>=t; i--) #define BUG puts("here!!!") #define STOP system("pause") #define file_r(x) freopen(x, "r", stdin) #define file_w(x) freopen(x, "w", stdout) using namespace std; const int MAX = 110; const double pi = acos(-1.0); struct point { double x, y; }; point p[MAX]; point c; double r; const double eps = 1e-8; int dcmp(double x) { return x < -eps ? -1 : x > eps ? 1 : 0; } double disp2p(point a,point b) // a b 两点之间的距离 { return sqrt( ( a.x - b.x ) * ( a.x - b.x ) + ( a.y - b.y ) * ( a.y - b.y ) ); } double crossProduct(point a,point b,point c)//向量 ac 在 ab 的方向 顺时针是正 { return (c.x - a.x)*(b.y - a.y) - (b.x - a.x)*(c.y - a.y); } point l2l_inst_p(point u1,point u2,point v1,point v2) { point ans = u1; double t = ((u1.x - v1.x)*(v1.y - v2.y) - (u1.y - v1.y)*(v1.x - v2.x))/ ((u1.x - u2.x)*(v1.y - v2.y) - (u1.y - u2.y)*(v1.x - v2.x)); ans.x += (u2.x - u1.x)*t; ans.y += (u2.y - u1.y)*t; return ans; } //直线与圆的交点 int l2c_inst_p(point c,double r,point l1,point l2,point *pv) { int cnt = 0; double d = fabs( crossProduct(c,l1,l2) )/disp2p(l1,l2); if( dcmp(d-r) > 0 ) return 0; point p = c; double t; p.x += l1.y - l2.y; p.y += l2.x - l1.x; p = l2l_inst_p(p,c,l1,l2); t = sqrt(r*r - disp2p(p,c)*disp2p(p,c))/disp2p(l1,l2); pv[cnt].x = p.x + (l2.x - l1.x)*t; pv[cnt++].y = p.y + (l2.y - l1.y)*t; if( dcmp(d-r) == 0 ) return cnt; pv[cnt].x = p.x - (l2.x - l1.x)*t; pv[cnt++].y = p.y - (l2.y - l1.y)*t; return cnt; } //若不包括端点，将等于号去掉 bool onSegment(point a, point b, point c) { //和端点重合 if( dcmp(c.x - a.x) == 0 && dcmp(c.y - a.y) == 0 || dcmp(c.x - b.x) == 0 && dcmp(c.y - b.y) == 0 ) return false; if( dcmp(crossProduct(a,b,c)) == 0 && dcmp(c.x - min(a.x,b.x)) >= 0 && dcmp(c.x-max(a.x,b.x)) <= 0 && dcmp(c.y - min(a.y,b.y)) >= 0 && dcmp(c.y-max(a.y,b.y)) <= 0 ) return true; return false; } //求线段与圆的交点,p中存放交点，返回交点个数 //若不包括端点，需要在onSegment函数中控制一下 int seg2c_inst_p(point c,double r,point l1,point l2,point *p) { point pv[3]; int cnt = l2c_inst_p(c, r, l1, l2, pv); int cntp = 0; FOR(i, 0, cnt) if( onSegment(l1, l2, pv[i]) ) p[cntp++] = pv[i]; //按照交点离l1的顺序从小到大排序 if( cntp == 2 ) { if(disp2p(p[0], l1) > disp2p(p[1], l1) ) swap(p[0], p[1]); } return cntp; } double area_triangle(point a,point b,point c) { return fabs( crossProduct(a,b,c) )/2.0; } //求以acb为圆心角的扇形（小于180度） double area_shan(point a, point b, point c, double r) { double aa = disp2p(c, b); double bb = disp2p(a, c); double cc = disp2p(a, b); double ang = acos((aa*aa + bb*bb - cc*cc)/(2*aa*bb)); return ang/2*r*r; } //三角形与圆的交面积，其中c是圆心 double area_triangle2circle(point a, point b, point c, double r) { point p[3]; int cnt = seg2c_inst_p(c, r, a, b, p); if( dcmp(crossProduct(a, b, c)) == 0 ) return 0; if( cnt == 0 ) { if( dcmp(disp2p(a, c) - r) <= 0 && dcmp(disp2p(b, c) - r) <= 0 ) return area_triangle(a, b, c); else return area_shan(a, b, c, r); } if( cnt == 1 ) { if( dcmp(disp2p(a, c) - r) > 0 && dcmp(disp2p(b, c) - r) > 0 ) return area_shan(a, b, c, r); if( dcmp(disp2p(a, c) - r) > 0 ) swap(a, b); return area_shan(p[0], b, c, r) + area_triangle(a, p[0], c); } if( cnt == 2 ) { return area_shan(a, p[0], c, r) + area_shan(p[1], b, c, r) + area_triangle(p[0], p[1], c); } } double solve(int n) { double area = 0; FOR(i, 0, n) { area += area_triangle2circle(p[i], p[(i+1)%n], c, r) * dcmp(crossProduct(c, p[(i+1)%n], p[i])); } return fabs(area); } void cal(double x0,double y0,double v0,double a,double t,double g) { double yt = v0*sin(a); double xt = v0*cos(a); double xx = x0 + xt*t; double yy = y0 + yt*t-0.5*g*t*t; c.x = xx; c.y = yy; } int main() { int n; double x, y, v, ang, t, g; while( ~scanf("%lf%lf%lf%lf%lf%lf%lf", &x, &y, &v, &ang, &t, &g, &r) ) { if( x == y && y == v && v == ang && ang == t && t == g && g == r && r == 0 ) break; scanf("%d", &n); FOR(i, 0, n) scanf("%lf%lf", &p[i].x, &p[i].y); cal(x, y, v, ang/180*pi, t, g); double ans = solve(n); printf("%.2lf\n", ans); } return 0; }
