P4196 [CQOI2006]凸多边形(半平面交)

传送门

题解：使用排序增量算法，先占个位置，这个代码是我测试过所有此模板中还未发现bug的代码.

附上代码：

#include<bits/stdc++.h>

using namespace std;

const double eps=1e-13;

struct point
{
    double x,y;
    point operator -(point &s){return (point){x-s.x,y-s.y};}
};

double operator *(point a,point b)
{
    return a.x*b.y-a.y*b.x;
}

struct line
{
    double d;
    point a,b;
}l[1005];

bool cmpd(line a,line b)
{
    return a.d<b.d;
}

//bool bian(point Q,point P1,point P2)
//{
//    return fabs((Q-P1)*(P2-P1)<eps&&min(P1.x,P2.x)-eps<=Q.x&&Q.x-eps<=max(P1.x,P2.x)&&min(P1.y,P2.y)-eps<=Q.y&&Q.y-eps<=max(P1.y,P2.y));
//}

point crosp(line a,line b)
{
    double s1=(b.a-a.a)*(a.b-a.a),s2=(a.b-a.a)*(b.b-a.a);
    return (point){(b.a.x*s2+s1*b.b.x)/(s1+s2),(b.a.y*s2+s1*b.b.y)/(s1+s2)};
}

int n,m,sta[2005];

int main()
{
    cin>>n;
    for (int i=1;i<=n;i++){
        int mm;
        cin>>mm;
        for (int j=1;j<=mm;j++){
            m++,cin>>l[m].a.x>>l[m].a.y;
            l[(j==1)?m+mm-1:m-1].b=l[m].a;
        }
    }
    n=m;
    for (int i=1;i<=n;i++){
        l[i].d=atan2(l[i].b.y-l[i].a.y,l[i].b.x-l[i].a.x);
    }
    sort(l+1,l+n+1,cmpd);
    for (int i=1;i<=n;i++){
        for (;sta[0]>=1;sta[0]--){
            if (fabs(l[i].d-l[sta[sta[0]]].d)<eps){
                if ((l[sta[sta[0]]].b-l[sta[sta[0]]].a)*(l[i].a-l[sta[sta[0]]].a)<eps){
                    break;
                }
            }else{
                break;
            }
        }
        for (;sta[0]>=2;sta[0]--){
            if ((l[i].b-l[i].a)*(crosp(l[sta[sta[0]]],l[sta[sta[0]-1]])-l[i].a)>eps){
                break;
            }
        }
        if (fabs(l[i].d-l[sta[sta[0]]].d)>=eps){
            sta[++sta[0]]=i;
        }
    }
    int L=1,R=sta[0];
    while (L<R){
        if ((l[sta[L]].b-l[sta[L]].a)*(crosp(l[sta[R]],l[sta[R-1]])-l[sta[L]].a)<eps){
           R--;
        }else{
            if ((l[sta[R]].b-l[sta[R]].a)*(crosp(l[sta[L]],l[sta[L+1]])-l[sta[R]].a)<eps){
               L++;
            }else{
               break;
            }
        }
    }
    if (R-L<=1){
        printf("0.000\n");
        return 0;
    }
    double ans=0;
    sta[R+1]=sta[L],sta[R+2]=sta[L+1];
    for (int i=L;i<=R;i++){
        ans+=crosp(l[sta[i]],l[sta[i+1]])*crosp(l[sta[i+1]],l[sta[i+2]])/2;
    }
    printf("%.3lf\n",ans);
    return 0;
}