• bzoj 2618: [Cqoi2006]凸多边形


    (我以为这道题要求出任意两个凸包的交,然后在容斥呢。。因为我竟然一开始以为万一凸包交出来的不是凸包怎么办。。)

    直接把所有边放在一起,半平面交就行。

     1 #include<cstdio>
     2 #include<algorithm>
     3 #include<cmath>
     4 #define eps 1e-8
     5 using namespace std;
     6 inline int ra()
     7 {
     8     int x=0,f=1; char ch=getchar();
     9     while (ch<'0' || ch>'9'){if (ch=='-') f=-1; ch=getchar();}
    10     while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
    11     return x*f;
    12 }
    13 int n,cnt,tot;
    14 double ans;
    15 struct point{double x,y;}p[505],a[505];
    16 struct line{point a,b; double angle;} l[505],s[505];
    17 point operator - (point a, point b){
    18     point t; t.x=a.x-b.x; t.y=a.y-b.y; return t;
    19 }
    20 double operator * (point a, point b){
    21     return a.x*b.y-a.y*b.x;
    22 }
    23 bool operator < (line a, line b){
    24     if (a.angle==b.angle) return (b.b-a.a)*(a.b-a.a)>0;
    25     return a.angle<b.angle;
    26 }
    27 point intersection(line a, line b)
    28 {
    29     double k1,k2,t; 
    30     k1=(b.a-a.a)*(a.b-a.a);
    31     k2=(a.b-a.a)*(b.b-a.a);
    32     t=k1/(k1+k2);
    33     point ans;
    34     ans.x=b.a.x+(b.b.x-b.a.x)*t;
    35     ans.y=b.a.y+(b.b.y-b.a.y)*t;
    36     return ans;
    37 }
    38 bool jud(line a, line b, line t)
    39 {
    40     point p=intersection(a,b);
    41     //printf("%lf %lf",p.x,p.y); while (1);
    42     return (t.a-p)*(t.b-p)<0;
    43 }
    44 void half_plane_intersection()
    45 {
    46     sort(l+1,l+cnt+1);
    47     int L=1,R=0;
    48     for (int i=1; i<=cnt; i++)
    49         if (l[i].angle!=l[i-1].angle) l[++tot]=l[i];
    50     cnt=tot; s[++R]=l[1]; s[++R]=l[2];
    51 //    for (int i=1; i<=cnt; i++)
    52 //        printf("%.1lf %.1lf %.1lf %.1lf
    ",l[i].a.x,l[i].a.y,l[i].b.x,l[i].b.y);
    53     for (int i=3; i<=cnt; i++)
    54     {
    55         while (L<R && jud(s[R],s[R-1],l[i])) R--;
    56         while (L<R && jud(s[L],s[L+1],l[i])) L++;
    57         s[++R]=l[i];
    58     }
    59     while (L<R && jud(s[R],s[R-1],s[L])) R--;
    60     while (L<R && jud(s[L],s[L+1],s[R])) L++;
    61     tot=0;
    62     s[1+R]=s[L];
    63     for (int i=L; i<=R; i++)
    64         a[++tot]=intersection(s[i],s[i+1]);
    65 }
    66 void get_ans()
    67 {
    68     if (tot<3) return;
    69     a[++tot]=a[1];
    70     for (int i=1; i<tot; i++)
    71         ans+=a[i]*a[i+1];
    72     ans=fabs(ans)/2;
    73 }
    74 int main()
    75 {
    76     int T=ra();
    77     while (T--)
    78     {
    79         n=ra();
    80         for (int i=1; i<=n; i++)
    81             p[i].x=ra(),p[i].y=ra();
    82         p[n+1]=p[1];
    83         for (int i=1; i<=n; i++)
    84             l[++cnt].a=p[i],l[cnt].b=p[i+1];
    85     }
    86     for (int i=1; i<=cnt; i++)
    87         l[i].angle=atan2(l[i].b.y-l[i].a.y,l[i].b.x-l[i].a.x);
    88     half_plane_intersection();
    89     get_ans();
    90     printf("%.3lf",ans);
    91     return 0;
    92 }
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  • 原文地址:https://www.cnblogs.com/ccd2333/p/6488646.html
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