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

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

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<cmath>
 6 #define inf 1e10
 7 #define maxn 10010
 8 using namespace std;
 9 int n,head,tail,tot,cnt;
10 double ans;
11 const double eps=1e-15;
12 struct point{double x,y;}p[maxn];
13 point operator +(point x,point y){return (point){x.x+y.x,x.y+y.y};}
14 point operator -(point x,point y){return (point){x.x-y.x,x.y-y.y};}
15 struct line{double ang,a,b,c;point pt;}li[maxn],que[maxn];
16 double dot(point a,point b){return a.x*b.x+a.y*b.y;}
17 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);}
18 bool includ(line x,point y){return y.x*x.a+y.y*x.b+x.c<=eps;}
19 bool comp(line x,line y){
20     if(x.ang==y.ang)return includ(y,x.pt);
21     return x.ang<y.ang;
22 }
23 point calc(line s1,line s2){
24     double v1=s1.b*s2.c-s1.c*s2.b,v2=s1.c*s2.a-s1.a*s2.c;
25     double v0=s1.a*s2.b-s1.b*s2.a;
26     return (point){v1/v0,v2/v0};
27 }
28 bool check(line x,line y,line z){return !includ(z,calc(x,y));}
29 bool solve(){
30     head=1;tail=0;
31     for(int i=1;i<=tot;i++){
32         if(i>1&&fabs(li[i].ang-li[i-1].ang)<=eps)continue;
33         while(head<tail&&check(que[tail-1],que[tail],li[i]))tail--;
34         while(head<tail&&check(que[head],que[head+1],li[i]))head++;
35         que[++tail]=li[i];
36     }
37     while(head<tail&&check(que[tail-1],que[tail],que[head]))tail--;
38     while(head<tail&&check(que[head],que[head+1],que[tail]))head++;
39     for(int i=head;i<tail;i++)p[++cnt]=calc(que[i],que[i+1]);
40     p[++cnt]=calc(que[head],que[tail]);
41     p[cnt+1]=p[1];
42 }
43 void getans(){
44     ans=0;for(int i=1;i<=cnt;i++)ans+=cross(p[i],p[i+1],(point){0,0});ans=fabs(ans)/2;
45 }
46 int main(){
47     //freopen("polygon.in","r",stdin);
48     //freopen("polygon.out","w",stdout); 
49     scanf("%d",&n);
50     for(int i=1;i<=n;i++){
51         double x,y,z;
52         scanf("%lf%lf%lf",&x,&y,&z);
53         li[++tot].a=x;li[tot].b=y;li[tot].c=z;
54     }
55     ++tot,li[tot].a=-1,li[tot].b=0,li[tot].c=-inf;
56     ++tot,li[tot].a=1,li[tot].b=0,li[tot].c=-inf;
57     ++tot,li[tot].a=0,li[tot].b=-1,li[tot].c=-inf;
58     ++tot,li[tot].a=0,li[tot].b=1,li[tot].c=-inf;
59     for(int i=1;i<=tot;i++){
60         li[i].ang=atan2(li[i].b,li[i].a);
61         if(li[i].b) li[i].pt=(point){0,-li[i].c/li[i].b};
62         else li[i].pt=(point){-li[i].c/li[i].a,0};
63     }
64     sort(li+1,li+tot+1,comp);
65     solve();getans();
66     printf("%.0lf
",ans);
67     return 0;
68 }

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