[POJ]2451: Uyuw's Concert

最新推荐文章于 2019-12-24 09:45:27 发布

转载最新推荐文章于 2019-12-24 09:45:27 发布 · 137 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/ditoly/p/POJ2451.html

本文介绍了一个关于计算有限制条件下舞台最大面积的问题解决方法。利用半平面交模板，通过具体的C++实现，详细展示了如何处理有向线段来确定舞台的最大可能区域。

题目大意：一个10000*10000的舞台，N条有向线，线左边才能布置舞台，求最后舞台面积。（N<=20000）

思路：半平面交模板。

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define MN 20000
#define eps 1e-7
struct point{double x,y;point(double x=0,double y=0):x(x),y(y){}}p[MN+5];
point operator+(point a,point b){return point(a.x+b.x,a.y+b.y);}
point operator-(point a,point b){return point(a.x-b.x,a.y-b.y);}
point operator*(point a,double x){return point(a.x*x,a.y*x);}
double operator*(point a,point b){return a.x*b.y-a.y*b.x;}
struct line{point p,v;line(double a=0,double b=0,double c=0,double d=0):p(a,b),v(c,d){}}l[MN+5];
bool cmp(line a,line b){return atan2(a.v.y,a.v.x)<atan2(b.v.y,b.v.x);}
bool left(line l,point p){return l.v*(p-l.p)<0;}
point xp(line a,line b){return a.p+a.v*(b.v*(a.p-b.p)/(a.v*b.v));}
int q[MN+5],ql,qr;
int main()
{
    int n,i;double ans=0;
    scanf("%d",&n);
    for(i=0;i<n;++i)
        scanf("%lf%lf%lf%lf",&l[i].p.x,&l[i].p.y,&l[i].v.x,&l[i].v.y),
        l[i].v.x-=l[i].p.x,l[i].v.y-=l[i].p.y;
    l[n++]=line(0,0,10000,0);
    l[n++]=line(10000,0,0,10000);
    l[n++]=line(10000,10000,-10000,0);
    l[n++]=line(0,10000,0,-10000);
    sort(l,l+n,cmp);
    for(i=1;i<n;++i)
    {
        while(ql<qr&&left(l[i],p[qr-1]))--qr;
        while(ql<qr&&left(l[i],p[ql]))++ql;
        if(q[++qr]=i,fabs(l[q[qr-1]].v*l[i].v)<eps)
            if(left(l[i],l[q[--qr]].p))q[qr]=i;
        if(ql<qr)p[qr-1]=xp(l[q[qr]],l[q[qr-1]]);
    }
    while(ql<qr&&left(l[q[ql]],p[qr-1]))--qr;
    p[qr]=xp(l[q[qr]],l[q[ql]]);
    for(i=ql;++i<qr;)ans+=(p[i]-p[ql])*(p[i+1]-p[ql]);
    printf("%.1f",ans/2);
}