fafu 1413

叉积的运用 ,不断的用叉积去判断 最小的拼图, 刚开始对点进行排序,每个人的排序规则不同做法可能不同,我是按照点的x轴进行x轴相同用y小的在前面,然后每个点按照最下的点开始进行查找 每次从一个点出发然后结束后无论找不找到都得 将出发的那条边删掉,ok然后就可以不断的去瓜分这张大的图,

#include <iostream>
#include <cstdio>
#include<string.h>
#include<cmath>
#include<algorithm>
#include<map>
#include<vector>
using namespace std;
struct point
{
    double x,y;
    point(double a=0,double b=0)
    {x=a;y=b;}
}node[1500],tong[1500];
struct line
{
    point p1,p2;
}L[1005];
point operator -(point a,point b){  return point(a.x-b.x,a.y-b.y);    }
bool operator <(const point &a,const point &b)
{ return a.x<b.x||(a.x==b.x&&a.y<b.y);}
bool operator ==(const point &a,const point &b){ return a.x==b.x&&a.y==b.y;    }
int n,sum,num[1500],ct;
double A[1500];
bool mark[1500];
int vis[1500];
bool cmp(point a,point b)
{return a.x<b.x||(a.x==b.x&&a.y<b.y);}
map<point,int>P;
vector<int>F[1500];
double area(int T)
{
    double sum=0;
    tong[T]=tong[0];
    for(int i=0;i<T;i++)
    sum+=tong[i].x*tong[i+1].y-tong[i].y*tong[i+1].x;
    if(sum>0) return sum/2;
    else return -sum/2;
}
void shifang(int a,int b)
{
	int i,ll,N;N=F[a].size();
    for(i=0;i<N;i++)
	{    ll=F[a][i];
	  if(ll==b) break;
	}
     F[a].erase(F[a].begin()+i);
	N=F[b].size();
	 for(i=0;i<N;i++)
     {
	     ll=F[b][i];
		 if(ll==a)
			 break;
	 }
	 F[b].erase(F[b].begin()+i);
}
double cross(point a,point b){return a.x*b.y-a.y*b.x;}
int look(int fa,int now,int nu)
{

	    if(vis[now]==2&&mark[now]) return nu;
     	tong[nu]=node[now];

      point gh[1000],p1=node[now],p2=node[now];
     int a=0,i,nut=0,T=0; bool k1,k2; k1=k2=false;
     if(vis[now]) a=1;
     for(i=0;i<F[now].size();i++)
	 {
		 int ll=F[now][i];
		 if((vis[ll]==0||a==0)&&mark[ll]&&ll!=fa)
        {

             gh[nut++]=node[ll];
        }
	 }
     for(i=0;i<nut;i++)
            if(cross(node[now]-node[fa],gh[i]-node[fa])>=0)
            {
				if(cross(gh[i]-node[now],p1-node[now])<0||p1==node[now])
				{	p1=gh[i];  k1=1;}
			}
			else
			{
				if(p2==node[now]||cross(p2-node[now],gh[i]-node[now])>0)
				{p2=gh[i]; k2=1; }
			}

     vis[now]=1;
	 if(!(p1==node[now])||k1)
	 {
          int ll=P[p1]  ;
	  	 T=look(now,ll,nu+1);
	 }
	 else if(!(p2==node[now])||k2){
		  int ll=P[p2];
		 T=look(now,ll,nu+1);
	 }
     vis[now]=0;
     return T;
}
point G;
bool cmp1(point a,point b)
{
	point ll=G;
    return cross(a-ll,b-ll)>=0;
}
bool cmp2(point a,point b)
{
	return a.y<b.y||(a.y==b.y&&a.x<=b.x);
}
void work()
{
  int k,j,i,T;
point pp[1500];
    for(i=0;i<sum;i++)
    {
		G=node[i];
		int a=P[node[i]];
        vis[a]=2;
          for(k=0;k<F[a].size();k++)
            pp[k]=node[F[a][k]];
		  sort(pp,pp+k,cmp2);
		sort(pp,pp+k,cmp1);
        for(j=0;j<k;j++)
        {
            tong[0]=node[a];
			int gg=P[pp[j]];
            if((T=look(a,gg,1)))
            {
               A[ct++]=area(T);
			  // for(int kk=0;kk<T;kk++)
				//   printf("%.3lf %.3lf    ",tong[kk].x,tong[kk ].y);
			   //printf("\n\n");
            }
            shifang(a,gg);
        }
		vis[a]=1;
        mark[a]=false;
	}
}
int main()
{
    point t1,t2;
    int i,a,b,f=0;
	//freopen("data.txt","r",stdin);
    while(scanf("%d",&n)==1)
    {

		if(f)puts("");f=1;
		 P.clear();
        ct=0;
        memset(mark,true,sizeof(mark));memset(num,0,sizeof(num));
        memset(vis,0,sizeof(vis));
        for(i=0;i<1500;i++)
           F[i].clear();
           sum=0;
       for(i=0;i<n;i++)
       {
           scanf("%lf%lf%lf%lf",&L[i].p1.x,&L[i].p1.y,&L[i].p2.x,&L[i].p2.y);
           a=P[L[i].p1];
           b=P[L[i].p2];
           if(a==0)
           {node[sum++]=L[i].p1; a=P[L[i].p1]=sum;}
           if(b==0)
           { node[sum++]=L[i].p2;b=P[L[i].p2]=sum;}
       }
       P.clear();
       sort(node,node+sum,cmp);
       for(i=0;i<sum;i++)
         P[node[i]]=i;
	   for(i=0;i<n;i++)
	   {
           a=P[L[i].p1];b=P[L[i].p2];
		   F[a].push_back(b);
		   F[b].push_back(a);
	   }
       work();
	   sort(A,A+ct);
	   for(i=0;i<ct;i++)
		   printf("%.2lf\n",A[i]);
	   P.clear();
    }
    return 0;
}