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("
");
}
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
",A[i]);
P.clear();
}
return 0;
}