BZOJ1502: [NOI2005]月下柠檬树

Simpson法相当好用啊!神奇的骗分算法!

 1 /**************************************************************
 2     Problem: 1502
 3     User: zhuohan123
 4     Language: C++
 5     Result: Accepted
 6     Time:228 ms
 7     Memory:1312 kb
 8 ****************************************************************/
 9  
10 #include <iostream>
11 #include <cstdio>
12 #include <cstring>
13 #include <cmath>
14 #include <algorithm>
15 using namespace std;
16 const double eps=1e-8,pi=3.141592653589793238;
17 int n;
18 struct point
19 {
20     double x,y;
21     point(){}
22     point(double X,double Y){x=X,y=Y;}
23 };
24 struct line
25 {
26     point s,e;
27     line(){}
28     line(point S,point E){s=S,e=E;}
29     double y(double x){return (e.y*s.x-e.x*s.y-e.y*x+s.y*x)/(s.x-e.x);}
30 }l[510];int lnum;
31 struct circle{double x,r;}c[510];
32 double h[510];
33 double f(double x)
34 {
35     double s=0;
36     for(int i=1;i<n;i++)
37     {
38         if(abs(x-c[i].x)+eps<c[i].r)s=max(s,sqrt(c[i].r*c[i].r-(x-c[i].x)*(x-c[i].x)));
39         if(l[i].s.x<x+eps&&l[i].e.x>x-eps)s=max(s,l[i].y(x));
40     }
41     return s*2;
42 }
43 const double dev=1e-6;
44 inline double simpson(double l,double r,double fl,double fm,double fr){return (fl+4*fm+fr)/6*(r-l);}
45 double integral(double l,double fl,double m,double fm,double r,double fr,double pre)
46 {
47     double lm=(l+m)/2,rm=(m+r)/2,flm=f(lm),frm=f(rm);
48     double intl=simpson(l,m,fl,flm,fm),intr=simpson(m,r,fm,frm,fr);
49     return abs(intl+intr-pre)<dev?intl+intr:integral(l,fl,lm,flm,m,fm,intl)+integral(m,fm,rm,frm,r,fr,intr);
50 }
51 int main(int argc, char *argv[])
52 {
53     double alp;scanf("%d%lf",&n,&alp);n++;
54     for(int i=1;i<=n;i++)scanf("%lf",&h[i]);
55     for(int i=1;i<n;i++)scanf("%lf",&c[i].r);c[n].r=0;
56     double s=1e10,e=-1e10;
57     for(int i=1;i<=n;i++)
58     {
59         h[i]+=h[i-1];
60         c[i].x=h[i]/tan(alp);
61         s=min(s,c[i].x-c[i].r);
62         e=max(e,c[i].x+c[i].r);
63     }
64     for(int i=1;i<n;i++)
65     {
66         double dx=c[i+1].x-c[i].x,dr=c[i].r-c[i+1].r;
67         if(abs(dx)<abs(dr)+eps)continue ;
68         l[++lnum]=line(point(c[i].x+c[i].r/dx*dr,sqrt(c[i].r*c[i].r-(c[i].r/dx*dr)*(c[i].r/dx*dr)))
69                       ,point(c[i+1].x+c[i+1].r/dx*dr,sqrt(c[i+1].r*c[i+1].r-(c[i+1].r/dx*dr)*(c[i+1].r/dx*dr))));
70     }
71     double m=(s+e)/2,fm=f(m);
72     printf("%.2lf
",integral(s,0,m,fm,e,0,simpson(s,e,0,fm,0)));
73     return 0;
74 }
原文地址:https://www.cnblogs.com/zhuohan123/p/3326475.html