poj 2074 Line of Sight

来源：互联网发布：电脑声控软件编辑：程序博客网时间：2024/05/16 23:02
有一个trick就是如果house和 property line之间没有障碍物直接输出property line的长度即可
#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<algorithm>using namespace std;struct P{    double x,y;    P (){}    P (double x,double y):x(x),y(y){    }    P operator -(P p){        return P(x-p.x,y-p.y);    }    P operator +(P p){        return P(x+p.x,y+p.y);    }    P operator *(double d){        return P(x*d,y*d);    }    double det(P p){        return (x*p.y)-(y*p.x);    }};struct Line{    P a,b;};Line H,p;Line qs[100000],jiaodian[100000],jiaodian1[100000];bool judge(P a,P b,P c,P d){    if((c-a).det(b-a)*(d-a).det(b-a)<=0)    return 1;    return 0;}P in(P a,P b,P c,P d){    return c+(d-c)*((b-a).det(a-c) / (b-a).det(d-c));}double dis(P a,P b){    double c=(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);    return sqrt(c);}bool cmp(Line a,Line b){    if(a.a.x!=b.a.x)    return a.a.x<b.a.x;    return a.b.x<b.b.x;}bool cmp1(Line a,Line b){return a.b.x<b.b.x;}int main(){    double x1,x2,y1;    while(~scanf("%lf%lf%lf",&x1,&x2,&y1))    {int i;        if(x1==0 && x2==0 && y1==0)        break;        H.a.x=x1;H.a.y=y1;        H.b.x=x2;H.b.y=y1;        scanf("%lf%lf%lf",&p.a.x,&p.b.x,&p.a.y);        p.b.y=p.a.y;        int k,l=0,l1=0,l2=0;        scanf("%d",&k);        while(k--)        {            scanf("%lf%lf%lf",&x1,&x2,&y1);if(x1>p.b.x || x2<p.a.x)continue;            if(y1>min(H.a.y,p.a.y) && y1<max(H.a.y,p.a.y))            {                qs[l].a.x=x1;                qs[l].b.x=x2;                qs[l].a.y=qs[l].b.y=y1;                l++;            }        }if(l==0){printf("%.2lf\n",p.b.x-p.a.x);continue;}        for( i=0;i<l;i++)        {P lf,rg;lf=in(H.b,qs[i].a,p.a,p.b);rg=in(H.a,qs[i].b,p.a,p.b);            if(judge(H.b,qs[i].a,p.a,p.b))            jiaodian[l1].a=in(H.b,qs[i].a,p.a,p.b);            else            {jiaodian[l1].a.x=p.a.x;jiaodian[l1].a.y=p.a.y;}            if(judge(H.a,qs[i].b,p.a,p.b))            jiaodian[l1].b=in(H.a,qs[i].b,p.a,p.b);            else            {jiaodian[l1].b.x=p.b.x;jiaodian[l1].b.y=p.b.y;}            l1++;if(lf.x>p.b.x || rg.x<p.a.x)l1--;//printf("%lf %lf\n",jiaodian[l1-1].a.x,jiaodian[l1-1].b.x);        }for(i=0;i<l1;i++){int fg=1;for(int j=0;j<l1;j++){if(i==j)continue;if(jiaodian[i].a.x>=jiaodian[j].a.x && jiaodian[i].b.x<=jiaodian[j].b.x)fg=0;}if(fg)jiaodian1[l2++]=jiaodian[i];}        sort(jiaodian1,jiaodian1+l2,cmp);        double ans=dis(p.a,jiaodian1[0].a);        for( i=0;i<l2-1;i++)        {            int fg=1;            Line hehe;            hehe.a.x=0x7ffffff;//printf("%lf %lf\n",jiaodian1[i].a.x,jiaodian1[i].b.x);            for(int j=i+1;j<l2;j++)            {                if(jiaodian1[j].a.x<=jiaodian1[i].b.x)                fg=0;                else                {                    if(jiaodian1[j].a.x<hehe.a.x)                        hehe=jiaodian1[j];                }            }            if(fg)            {                double ans1=dis(jiaodian1[i].b,hehe.a);                if(ans1>ans)                ans=ans1;            }        }sort(jiaodian1,jiaodian1+l2,cmp1);        double llll=dis(jiaodian1[l2-1].b,p.b);        if(llll>ans)        ans=llll;        if(ans!=0)        printf("%.2lf\n",ans);        else        printf("No View\n");    }    return 0;}
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