• 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 }
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  • 原文地址:https://www.cnblogs.com/zhuohan123/p/3326475.html
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