「SDOI2018」物理实验

题目大意:

  这题不好描述,直接看原题吧……

题解:

  很无脑的题……就是卡精度+难写。代码能力还是太差了。

  其实可以直接用long double肝过去。但我的代码似乎太丑了,以至于跑得奇慢无比。

代码:

  

#include "bits/stdc++.h"

using namespace std;


inline int read() {
    int s=0,k=1;char ch=getchar();
    while (ch<'0'|ch>'9') ch=='-'?k=-1:0,ch=getchar();
    while (ch>47&ch<='9') s=s*10+(ch^48),ch=getchar();
    return s*k;
}

#define double long double 

const double eps=1e-10;

const int N=1e4+10;

inline int dcmp(double a) {
    if (a>eps) return 1;
    if (a<-eps) return -1;
    return 0; 
}

struct P{
    P(double _x=0.0,double _y=0.0):x(_x),y(_y){}
    double x,y;
    double &operator [] (int k) {return k?y:x;}
    friend bool operator <(P a,P b) {
        return dcmp(a[0]-b[0])==-1||(dcmp(a[0]-b[0])==0&&dcmp(a[1]-b[1])==-1);
    }
    friend double operator *(P a,P b) {
        return a[0]*b[1]-b[0]*a[1];
    }
    friend P operator /(P a,double b) {
        return P(a[0]/b,a[1]/b);
    }
    friend P operator + (P a,P b) {
        return P(a[0]+b[0],a[1]+b[1]);
    }
    friend P operator * (P a,double b) {
        return P(a[0]*b,a[1]*b);
    }
    friend double operator /(P a,P b) {
        return a[0]*b[0]+b[1]*a[1];
    }
    inline double leth(){
        return sqrt(x*x+y*y);
    }
};
P operator -(P a,P b) {
    return P(a[0]-b[0],a[1]-b[1]);
}
    
struct L{
    L(){}L(P _a,P _b):a(_a),b(_b),c(_b-_a),slope(atan2((a-b)[1],(a-b)[0])){}
    P a,b,c;double slope;
    P &operator[] (int k) {
        if(!k) return a;
        if (k==1) return b;
        return c;
    }
    friend double touying (L x,L y) {
        return x[2]/y[2]/(x[2].leth());
    }
    friend double touying (L x,P y) {
        return x[2]/(y-x[0])/(x[2].leth());
    }
    friend bool operator <(L x,L y) {
        double pos = max (x[0][0],y[0][0]);
        return dcmp(fabs(x[0][1] + x[2][1] * (pos-x[0][0]) / ( x[1][0] - x[0][0] )) -
                   fabs(y[0][1] + y[2][1] * (pos-y[0][0]) / ( y[1][0] - y[0][0] )) ) == -1; 
                 
    }
}block[N],gui;

inline bool Judge(L a,L b) {
    return dcmp(b[2]*(a[0]-b[0]))==1;
}

inline double dis(P p,L l) {
    return l[2]*(p-l[0])/l[2].leth();
}

double lft[N],lth[N],rght[N],rth[N];

int n,leth,m;
int pl[N],pr[N];

struct node {
    P pos;int id;
    friend bool operator <(node a,node b) {
        return a.pos<b.pos;
    }
}np[N*6];

set<L> up,down;

double res[N*6];

inline double calc (double x2,double x1,L l) {

    P delta=l[2]*(x2-x1)/l[2][0];
    return delta.leth();
}

int main(){

    int T=read();

    register int i,j;
    while (T-- ) {
        n=read();
        for (i=1;i<=n;++i) {
            double x1=read(),y1=read(),x2=read(),y2=read();
            if (x1>x2) swap(x1,x2),swap(y1,y2);
            if (x1==x2&&y1>y2) swap(y1,y2);
            block[i]=L(P(x1,y1),P(x2,y2));
        }
        double x1=read(),y1=read(),x2=read(),y2=read();leth=read();
        if (x1>x2) swap(x1,x2),swap(y1,y2);
        if (x1==x2&&y1>y2) swap(y1,y2);
        gui=L(P(x1,y1),P(x2,y2));
        m=0;
        for (i=1;i<=n;++i) {
            P a=block[i][0],b=block[i][1];
            P na,nb;
            na[0] = touying(gui,a),na[1] = dis(a,gui);
            nb[0] = touying(gui,b),nb[1] = dis(b,gui);
            if (nb<na) swap(na,nb);
            block[i] = L(na,nb);
            np[++m] = (node){na,i} ;
            np[++m] = (node){nb,-i} ;
            np[++m] = (node){P (na[0]+leth,na[1]),0};
            np[++m] = (node){P (nb[0]+leth,nb[1]),0};
            np[++m] = (node){P (na[0]-leth,na[1]),0};
            np[++m] = (node){P (nb[0]-leth,nb[1]),0};
        }
        sort(np+1,np+m+1);    
        up.clear();
        down.clear();
        double ans = 0.0,tmp,ret=0.0;
        j=1;
        while (j<=6*n) {
            if (np[j].id<0) {
                if (block[-np[j].id][0][1]<-eps) 
                    down.erase(block[-np[j].id]);
                else up.erase(block[-np[j].id]);
            }else if (np[j].id>0){
                if (block[np[j].id][0][1]<-eps) 
                    down.insert(block[np[j].id]);
                else up.insert(block[np[j].id]);
            }
            ++j;
            res[j]=0.0;
            if (up.size()){
                tmp = calc (np[j].pos[0],np[j-1].pos[0],*up.begin());
                res[j] +=tmp;
            }
            if (down.size()) {
                tmp = calc (np[j].pos[0],np[j-1].pos[0],*down.begin());
                res[j] +=tmp;
            }
        }
        for (i=j=1;i<=6*n&&j<=6*n;++i,ans-=res[i]) {
            while (j<=6*n&&dcmp(np[j].pos[0]-np[i].pos[0]-leth)<=0) ans+=res[j++];
            ret=max(ret,ans);
        }
        printf("%.15Lf\n",ret);
    }
}

  

posted @ 2018-05-19 20:54  Troywar  阅读(496)  评论(0编辑  收藏