bzoj1502 simpson求面积

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=1502

题解：

simpson积分求面积，s = (f(a)+f(b)+4*f(c))/6*Δx ，c=(a+b)/2。

题中的树投影下来是一些圆和相邻圆的公切线组成的一个封闭图形，并且上下对称，所以可以只求上半部分。

simpson求面积时，若f(x)代价很大，要尽量减少其的重复调用。其次尽量优化f(x)函数的计算。

写完后还要自己出极限随机数据，将eps调到能接受的最大值。

 #include <cstdio>
 #include <cmath>
 #include <iostream>
 #define maxn 510
 #define eps 1e-6
 using namespace std;

 int sg( double x ) {
     return (x>-eps)-(x<eps);
 }
 struct Vector {
     double x, y;
     Vector(){}
     Vector( double x, double y ) : x(x), y(y) {}
     Vector operator+( Vector b ) { return Vector(x+b.x,y+b.y); }
     Vector operator-( Vector b ) { return Vector(x-b.x,y-b.y); }
     Vector operator*( double b ) { return Vector(x*b,y*b); }
     double len() {
         return sqrt( x*x+y*y );
     }
 };
 typedef Vector Point;

 struct Line {
     Point A, B;
     double k, b;
     double lf, rg;
     Line(){}
     Line( Point A, Point B ):A(A),B(B) {
         if( A.x>B.x ) swap(A,B);
         lf = A.x;
         rg = B.x;
         k = (B.y-A.y)/(B.x-A.x);
         b = A.y-A.x*k;
     }
     inline bool in( double x ) {
         return sg(x-lf)>= && sg(rg-x)>=;
     }
     inline double f( double x ) { return k*x+b; }
     void out() {
         printf( "(%lf,%lf) (%lf,%lf) \n", A.x, A.y, B.x, B.y );
     }
 };
 struct Circle {
     Point C;
     double r;
     double lf, rg;
     Circle(){}
     void make() {
         lf = C.x-r;
         rg = C.x+r;
     }
     inline bool in( double x ) {
         return sg(x-lf)>= && sg(rg-x)>=;
     }
     inline double f( double x ) {
         return sqrt( r*r-(C.x-x)*(C.x-x) );
     }
     void out() {
         printf( "(%lf,%lf) r=%lf\n", C.x, C.y, r );
     }
 };

 int n;
 double alpha;
 Line lines[maxn];
 Circle cirs[maxn];

 Line cir_tang( Circle & a, Circle & b ) {
     double ang = acos( (a.r-b.r)/(b.C.x-a.C.x) );
     Point A = a.C + Vector(cos(ang),sin(ang))*a.r;
     Point B = b.C + Vector(cos(ang),sin(ang))*b.r;
     return Line(A,B);
 }

 double F( double x ) {
     double y = -1.0;
     for( int i=; i<=n; i++ ) {
         if( cirs[i].lf-eps<=x && x<=cirs[i].rg+eps )
             y = max( y, cirs[i].f(x) );
         if( i&& lines[i].lf-eps<=x && x<=lines[i].rg+eps )
             y = max( y, lines[i].f(x) );
     }
     return y;
 }
 inline double simpson( double a, double b, double fa, double fb, double fc ) {
     return (fa+*fc+fb)*(b-a)/;
 }
 double adapt( double a, double b, double c, double fa, double fb, double fc ) {
     double d = (a+c)/, e = (c+b)/;
     double fd = F(d), fe = F(e);
     double sa = simpson(a,c,fa,fc,fd);
     double sb = simpson(c,b,fc,fb,fe);
     double ss = simpson(a,b,fa,fb,fc);
     if( fabs(sa+sb-ss)<*eps ) return sa+sb;
     return adapt(a,c,d,fa,fc,fd)+adapt(c,b,e,fc,fb,fe);
 }

 int main() {
     scanf( "%d%lf", &n, &alpha );
     double k = /tan(alpha);
     double hsum = 0.0, h;
     double lf=1e200, rg=-1e200;
     for( int i=; i<=n; i++ ) {
         scanf( "%lf", &h );
         hsum += h;
         cirs[i].C.x = hsum*k;
         cirs[i].C.y = 0.0;
     }
     for( int i=; i<n; i++ )  {
         scanf( "%lf", &cirs[i].r );
         cirs[i].make();
         lf = min( lf, cirs[i].lf );
         rg = max( rg, cirs[i].rg );
     }
     cirs[n].r = 0.0;
     rg = max( rg, cirs[n].C.x );
     for( int i=; i<=n; i++ )
         lines[i] = cir_tang( cirs[i-], cirs[i] );
     printf( "%.2lf\n", adapt(lf,rg,(lf+rg)/,F(lf),F(rg),F((lf+rg)/) )* );
 }