bzoj1845: [Cqoi2005] 三角形面积并

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传送门
裸的扫描线。
枚举两条线段。判断是否有交点
然后按照交点横坐标排序
然后就是线段覆盖个问题了。
#include<cmath>#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<algorithm>#define N 333#define ld long double#define pa pair<ld,ld>using namespace std;const ld eps=1e-8,inf=1e10;struct P{    ld x,y;    P(){}    P(ld _,ld __):x(_),y(__){}    void read(){        double _,__;        scanf("%lf%lf",&_,&__);        x=_,y=__;    }    friend bool operator <(P a,P b){        if (fabs(a.x-b.x)<eps) return a.y<b.y;        return a.x<b.x;    }    friend P operator +(P a,P b){        return P(a.x+b.x,a.y+b.y);    }    friend P operator -(P a,P b){        return P(a.x-b.x,a.y-b.y);    }    friend P operator *(ld a,P b){        return P(a*b.x,a*b.y);    }    friend ld operator *(P a,P b){        return a.x*b.x+a.y*b.y;    }    friend ld operator ^(P a,P b){        return a.x*b.y-b.x*a.y;    }}a[N][3],p[N*N];struct L{    P a,b;    L(){}    L(P _,P __){        a=_,b=__-_;    }    P operator [](int k){        return k?a+b:a;    }    friend bool cross(L x,L y){        return (x.b^(y[0]-x.a))*(x.b^(y[1]-x.a))<-eps&&(y.b^(x[0]-y.a))*(y.b^(x[1]-y.a))<-eps;    }    friend P getP(L x,L y){        P t=x.a-y.a;        ld tmp=(y.b^t)/(x.b^y.b);        return x.a+tmp*x.b;    }}l[N][3],T;pa f[N];int n,m,cnt,tot;ld ans,la,A,B,sum;int main(){    scanf("%d",&n);    for (int i=1;i<=n;i++){        a[i][0].read(),a[i][1].read(),a[i][2].read();        p[++tot]=a[i][0],p[++tot]=a[i][1],p[++tot]=a[i][2];        sort(a[i],a[i]+3);        if (((a[i][2]-a[i][0])^(a[i][1]-a[i][0]))>eps){            l[i][0]=L(a[i][0],a[i][2]);            l[i][1]=L(a[i][2],a[i][1]);            l[i][2]=L(a[i][1],a[i][0]);        }        else{            l[i][0]=L(a[i][2],a[i][0]);            l[i][1]=L(a[i][1],a[i][2]);            l[i][2]=L(a[i][0],a[i][1]);        }    }    for (int i=1;i<=n;i++)        for (int j=1;j<i;j++)            for (int x=0;x<3;x++)                for (int y=0;y<3;y++)                    if (cross(l[i][x],l[j][y]))                        p[++tot]=getP(l[i][x],l[j][y]);    sort(p+1,p+tot+1);    ans=0; la=p[1].x;    T=L(P(0,-inf),P(0,inf));    for (int i=2;i<=tot;i++){        T.a.x=(la+p[i].x)/2; cnt=0;        for (int j=1;j<=n;j++)            if (cross(l[j][0],T)){                A=getP(l[j][0],T).y;                if (cross(l[j][1],T))                    B=getP(l[j][1],T).y;                else B=getP(l[j][2],T).y;                if (A>B) swap(A,B);                f[++cnt]=pa(A,B);            }        sort(f+1,f+cnt+1);        sum=0; A=-inf;        for (int j=1;j<=cnt;j++){            if (f[j].first>A){                sum+=f[j].second-f[j].first;                A=f[j].second;            }            else if (f[j].second>A){                sum+=f[j].second-A;                A=f[j].second;            }        }        ans+=(p[i].x-la)*sum;        la=p[i].x;    }    printf("%.2lf",(double)ans);}
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