老oj1965：polygon半平面交

题目链接：http://192.168.2.240:8080/JudgeOnline/showproblem?problem_id=1965

polygon半平面交

Time Limit:1000MS  Memory Limit:165536K
Total Submit:66 Accepted:25 
Case Time Limit:100MS

Description

Input

n为半平面个数，以下n行表示若干半平面。形式如ax+by+c<=0

Output

输出半平面交的面积。保留3位小数 
保证面积有限。 
有多组数据：

Sample Input

4
1 0 -1
0 1 -1
-1 0 -1
0 -1 -1

Sample Output

4.000
数据保证：N<=10000
半平面交出的多边形的坐标在10^10以内。

Source

计算几何 半平面交

半平面交裸题。

讲道理不是说好保留三位小数吗，TMD数据全是保留到整数……幸好我WA了一次之后看了看数据，不然不知道要改多久……机智如我

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #include<cmath>
 #define inf 1e10
 #define maxn 10010
 using namespace std;
 int n,head,tail,tot,cnt;
 double ans;
 const double eps=1e-;
 struct point{double x,y;}p[maxn];
 point operator +(point x,point y){return (point){x.x+y.x,x.y+y.y};}
 point operator -(point x,point y){return (point){x.x-y.x,x.y-y.y};}
 struct line{double ang,a,b,c;point pt;}li[maxn],que[maxn];
 double dot(point a,point b){return a.x*b.x+a.y*b.y;}
 double cross(point x,point y,point z){return (x.x-z.x)*(y.y-z.y)-(x.y-z.y)*(y.x-z.x);}
 bool includ(line x,point y){return y.x*x.a+y.y*x.b+x.c<=eps;}
 bool comp(line x,line y){
     if(x.ang==y.ang)return includ(y,x.pt);
     return x.ang<y.ang;
 }
 point calc(line s1,line s2){
     double v1=s1.b*s2.c-s1.c*s2.b,v2=s1.c*s2.a-s1.a*s2.c;
     double v0=s1.a*s2.b-s1.b*s2.a;
     return (point){v1/v0,v2/v0};
 }
 bool check(line x,line y,line z){return !includ(z,calc(x,y));}
 bool solve(){
     head=;tail=;
     for(int i=;i<=tot;i++){
         if(i>&&fabs(li[i].ang-li[i-].ang)<=eps)continue;
         while(head<tail&&check(que[tail-],que[tail],li[i]))tail--;
         while(head<tail&&check(que[head],que[head+],li[i]))head++;
         que[++tail]=li[i];
     }
     while(head<tail&&check(que[tail-],que[tail],que[head]))tail--;
     while(head<tail&&check(que[head],que[head+],que[tail]))head++;
     for(int i=head;i<tail;i++)p[++cnt]=calc(que[i],que[i+]);
     p[++cnt]=calc(que[head],que[tail]);
     p[cnt+]=p[];
 }
 void getans(){
     ans=;for(int i=;i<=cnt;i++)ans+=cross(p[i],p[i+],(point){,});ans=fabs(ans)/;
 }
 int main(){
     //freopen("polygon.in","r",stdin);
     //freopen("polygon.out","w",stdout);
     scanf("%d",&n);
     for(int i=;i<=n;i++){
         double x,y,z;
         scanf("%lf%lf%lf",&x,&y,&z);
         li[++tot].a=x;li[tot].b=y;li[tot].c=z;
     }
     ++tot,li[tot].a=-,li[tot].b=,li[tot].c=-inf;
     ++tot,li[tot].a=,li[tot].b=,li[tot].c=-inf;
     ++tot,li[tot].a=,li[tot].b=-,li[tot].c=-inf;
     ++tot,li[tot].a=,li[tot].b=,li[tot].c=-inf;
     for(int i=;i<=tot;i++){
         li[i].ang=atan2(li[i].b,li[i].a);
         if(li[i].b) li[i].pt=(point){,-li[i].c/li[i].b};
         else li[i].pt=(point){-li[i].c/li[i].a,};
     }
     sort(li+,li+tot+,comp);
     solve();getans();
     printf("%.0lf\n",ans);
     return ;
 }