BZOJ1074: [SCOI2007]折纸origami

来源：互联网发布：python黑帽子和灰帽子编辑：程序博客网时间：2024/05/16 01:54

很裸的一道计算几何对吧
每一个询问我们就只需要易操作将他展开就好了。。
很显然爆搜中可以删几个不可能的中途状态点
时间上界是 2n∗m的
然后我是在linux打的。。。调戏了很久毛了搬到Windows上才发现是一个double打成了int。。。。。各种不爽

#include<cstdio>#include<iostream>#include<cstdio>#include<set>using namespace std;inline double abs(double a){return a<0?-a:a;}const   double eps=0.000001;struct Point{   double x,y;   inline friend bool operator ==(Point a,Point b) {          return abs(a.x-b.x)<eps&&abs(a.y-b.y)<eps; } };const   int INF=1<<29;const Point Wa=(Point){-INF,-INF};const Point Single=(Point){INF,INF};struct Line{   Point start,end;};inline double Cross(Point a,Point b){return a.x*b.y-a.y*b.x;}inline Point Rotato(Point a,Line b){      Point tpa,tpb;     tpa.x=a.x-b.start.x;     tpa.y=a.y-b.start.y;     tpb.x=b.end.x-b.start.x;     tpb.y=b.end.y-b.start.y;       Point res;     if(Cross(tpb,tpa)>eps)         {           if(abs(b.start.x-b.end.x)<eps)                    res.x=2*b.start.x-a.x,res.y=a.y;            else if(abs(b.start.y-b.end.y)<eps)                    res.x=a.x,res.y=2*b.start.y-a.y;            else                 {                    double k=b.end.y-b.start.y,k1;                    k/=b.end.x-b.start.x;                    k1=-1/k;double b1=b.end.x*b.start.y-b.end.y*b.start.x,b2;                    b1/=b.end.x-b.start.x;                    if(abs(a.x)<eps)b2=a.y;                    else b2=a.y-k1*a.x;                    res.x=(b2-b1)/(k-k1);res.y=k*res.x+b1;                    res.x*=2,res.y*=2,res.x-=a.x,res.y-=a.y;                  }            return res;          }    if(Cross(tpb,tpa)<-eps)    return Wa;    return Wa;}Line Oper[10001];int ans;int n;inline bool Can(Point a){return a.x>eps&&a.x<100-eps&&a.y>eps&&a.y<100-eps;}void DFS(Point a,int t){       Point fl;      if(t==0)        {if(Can(a))ans++;return ;}      fl=Rotato(a,Oper[t])     ;      if(fl==Wa)return;      else if(fl==Single)DFS(a,t-1);      else DFS(fl,t-1),DFS(a,t-1);}int main(){  int n,m;scanf("%d",&n);  for(int i=1;i<=n;i++)      scanf("%lf",&Oper[i].start.x),     scanf("%lf",&Oper[i].start.y),     scanf("%lf",&Oper[i].end.x),     scanf("%lf",&Oper[i].end.y);  scanf("%d",&m);  for(int i=1;i<=m;i++)        {            Point re;ans=0;           scanf("%lf%lf",&re.x,&re.y);           DFS(re,n);           printf("%d\n",ans);         }  return 0;}

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