Poj 2074 Line of Sight

地址：http://poj.org/problem?id=2074

题目：

Line of Sight

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 4148		Accepted: 1291

Description

An architect is very proud of his new home and wants to be sure it can be seen by people passing by his property line along the street. The property contains various trees, shrubs, hedges, and other obstructions that may block the view. For the purpose of this problem, model the house, property line, and obstructions as straight lines parallel to the x axis: 

To satisfy the architect's need to know how visible the house is, you must write a program that accepts as input the locations of the house, property line, and surrounding obstructions and calculates the longest continuous portion of the property line from which the entire house can be seen, with no part blocked by any obstruction.

Input

Because each object is a line, it is represented in the input file with a left and right x coordinate followed by a single y coordinate: 
< x1 > < x2 > < y > 
Where x1, x2, and y are non-negative real numbers. x1 < x2 
An input file can describe the architecture and landscape of multiple houses. For each house, the first line will have the coordinates of the house. The second line will contain the coordinates of the property line. The third line will have a single integer that represents the number of obstructions, and the following lines will have the coordinates of the obstructions, one per line. 
Following the final house, a line "0 0 0" will end the file. 
For each house, the house will be above the property line (house y > property line y). No obstruction will overlap with the house or property line, e.g. if obstacle y = house y, you are guaranteed the entire range obstacle[x1, x2] does not intersect with house[x1, x2].

Output

For each house, your program should print a line containing the length of the longest continuous segment of the property line from which the entire house can be to a precision of 2 decimal places. If there is no section of the property line where the entire house can be seen, print "No View".

Sample Input

2 6 6
0 15 0
3
1 2 1
3 4 1
12 13 1
1 5 5
0 10 0
1
0 15 1
0 0 0

Sample Output

8.80
No View
思路：首先排除掉不在house和proprety line的y的区间中线段，再求出对于每个线段的在property上的不可见区间。
　　如图所示：

  

然后把所有的区间按左端点为第一关键字，右端点为第二关键字从小到大排序，扫描一遍就好了。

/* 二维几何  */
/* 需要包含的头文件 */
#include<cstdio>
#include <cstring>
#include <cmath >
#include <iostream>
#include <algorithm>

using namespace std;
/** 常用的常量定义 **/
const double INF  = 1e200;
const double eps  = 1e-10;
const double PI  = acos(-1.0);
const int Max = 1e5;
/** 基本几何结构 **/
struct Point
{
double x,y;
Point(double a=0, double b=0){x=a,y=b;}
bool operator<(const Point &ta)const
{
if(x==ta.x)     return y<ta.y;
return x<ta.x;
}
friend Point operator+(const Point &ta,const Point &tb)
{
return Point(ta.x+tb.x,ta.y+tb.y);
}
friend Point operator-(const Point &ta,const Point &tb)
{
return Point(ta.x-tb.x,ta.y-tb.y);
}
};
struct Vec2D        ///二维向量，*重载为点乘，/重载为叉乘
{
double x,y;
Vec2D(double ta,double tb){x=ta,y=tb;}
Vec2D(Point &ta){x=ta.x,y=ta.y;}
friend double operator*(const Vec2D &ta,const Vec2D &tb)
{
return ta.x*tb.x+ta.y*tb.y;
}
friend double operator/(const Vec2D &ta,const Vec2D &tb)
{
return ta.x*tb.y-ta.y*tb.x;
}
friend Vec2D operator+(const Vec2D &ta,const Vec2D &tb)
{
return Vec2D(ta.x+tb.x,ta.y+tb.y);
}
friend Vec2D operator-(const Vec2D &ta,const Vec2D &tb)
{
return Vec2D(ta.x-tb.x,ta.y-tb.y);
}
Vec2D operator=(const Vec2D &ta)
{
x=ta.x,y=ta.y;
return *this;
}
};
struct LineSeg      ///线段，重载了/作为叉乘运算符，*作为点乘运算符
{
Point s,e;
LineSeg(){s=Point(0,0),e=Point(0,0);}
LineSeg(Point a, Point b){s=a,e=b;}
double lenth(void)
{
return sqrt((s.x-e.x)*(s.x-e.x)+(s.y-e.y)*(s.y-e.y));
}
friend double operator*(const LineSeg &ta,const LineSeg &tb)
{
return (ta.e.x-ta.s.x)*(tb.e.x-tb.s.x)+(ta.e.y-ta.s.y)*(tb.e.y-tb.s.y);
}
friend double operator/(const LineSeg &ta,const LineSeg &tb)
{
return (ta.e.x-ta.s.x)*(tb.e.y-tb.s.y)-(ta.e.y-ta.s.y)*(tb.e.x-tb.s.x);
}
LineSeg operator=(const LineSeg &ta)
{
s=ta.s,e=ta.e;
return *this;
}
};
struct Line         /// 直线的解析方程 a*x+b*y+c=0  为统一表示，约定 a >= 0
{
double a,b,c;
Line(double d1=1, double d2=-1, double d3=0){ a=d1,b=d2,c=d3;}
};

int sgn(double ta,double tb);
double fArea(Point &ta,Point &tb,Point &tc);
bool intersect(LineSeg &lx,LineSeg &ly);
bool intersection(LineSeg &lx,LineSeg &ly,Point &pt);
double getdis(const Point &ta,const Point &tb);
bool cmp(const Point &ta,const Point &tb);
void graham(Point ps[],Point tb[],int n,int &num);
void ConvexClosure(Point ps[],Point tb[],int n,int &num);

void scf(LineSeg &lx)
{
cin>>lx.s.x>>lx.e.x>>lx.s.y;
lx.e.y=lx.s.y;
}
LineSeg hs,pl,cur,lx;
Point line[Max],tx,ty;
int main(void)
{
while(1)
{
int n,num=0;
scf(hs);
if(!(hs.s.x||hs.s.y||hs.e.x))
break;
scf(pl);
cin>>n;
for(int i=0;i<n;i++)
{
scf(cur);
if(sgn(cur.s.y,hs.s.y)<0&&sgn(cur.s.y,pl.s.y)>0)
{
lx=LineSeg(hs.s,cur.e);
intersection(lx,pl,tx);
lx=LineSeg(hs.e,cur.s);
intersection(lx,pl,ty);
if(tx.x>=ty.x)
line[num++]=Point(ty.x,tx.x);
else
line[num++]=Point(tx.x,ty.x);
}
}
sort(line,line+num);
double ans=0,rr=pl.s.x;
line[num++]=Point(pl.e.x,pl.e.x);
for(int i=0;i<num;i++)
if(!(line[i].y<pl.s.x || line[i].x>pl.e.x))
{
//printf("====%f %f\n",line[i].x,line[i].y);
line[i].x=max(pl.s.x,line[i].x);
line[i].y=min(pl.e.x,line[i].y);
if(line[i].x>rr)
ans=max(ans,line[i].x-rr);
rr=max(line[i].y,rr);
}
if(sgn(ans,0))
printf("%.2f\n",ans);
else
printf("No View\n");

}

return 0;
}

/*******判断ta与tb的大小关系*******/
int sgn(double ta,double tb)
{
if(fabs(ta-tb)<eps)return 0;
if(ta<tb)   return -1;
return 1;
}
/*********求两点的距离*************/
double getdis(const Point &ta,const Point &tb)
{
return sqrt((ta.x-tb.x)*(ta.x-tb.x)+(ta.y-tb.y)*(ta.y-tb.y));
}
/************三角形面积**************************/
double fArea(Point &ta,Point &tb,Point &tc)
{
return fabs(LineSeg(ta,tb)/LineSeg(ta,tc)*0.5);
}

/*********** 判断P1P2是否和P3P4相交****************************
其中Pi坐标为(xi,yi),需要满足两个条件:
（1）快速排斥试验：
以P1P2为对角线的矩形S1是否和以P3P4为对角线的矩形S2相交，
即 min(x1,x2)<=max(x3,x4) && min(x3,x4)<=max(x1,x2)
&& min(y1,y2)<=max(y3,y4) &&min(y3,y4)<=max(y1,y2)
（2）跨立试验：
点P1，P2必然在线段P3P4的不同侧，
点P3，P4必然在线段P1P2的不同侧，
***************************************************************/
bool intersect(LineSeg &lx,LineSeg &ly)
{
return     sgn(min(lx.s.x,lx.e.x),max(ly.s.x,ly.e.x))<=0
&& sgn(min(ly.s.x,ly.e.x),max(lx.s.x,lx.e.x))<=0
&& sgn(min(lx.s.y,lx.e.y),max(ly.s.y,ly.e.y))<=0
&& sgn(min(ly.s.y,ly.e.y),max(lx.s.y,lx.e.y))<=0
&& sgn((lx/LineSeg(lx.s,ly.s))*(lx/LineSeg(lx.s,ly.e)),0)<=0
&& sgn((ly/LineSeg(ly.s,lx.s))*(ly/LineSeg(ly.s,lx.e)),0)<=0;
}
/************线段求交点**************************
返回-1代表直线平行，返回0代表直线重合，返回1代表线段相交
利用叉积求得点P分线段DC的比，
然后利用高中学习的定比分点坐标公式求得分点P的坐标
**************************************************/
bool intersection(LineSeg &lx,LineSeg &ly,Point &pt)
{
pt=lx.s;
if(sgn(lx/ly,0)==0)
{
if(sgn(LineSeg(lx.s,ly.e)/ly,0)==0)
return 0;//重合
return -1;//平行
}
double t = (LineSeg(lx.s,ly.s)/ly)/(lx/ly);
pt.x+=(lx.e.x-lx.s.x)*t, pt.y+=(lx.e.y-lx.s.y)*t;
return 1;
}
/** ************凸包算法****************
寻找凸包的graham 扫描法
PS(PointSet)为输入的点集；
tb为输出的凸包上的点集，按照逆时针方向排列;
n为PointSet中的点的数目
num为输出的凸包上的点的个数
****************************************** **/
bool cmp(const Point &ta,const Point &tb)/// 选取与最后一条确定边夹角最小的点，即余弦值最大者
{
//    double tmp=LineSeg(ps[0],ta)/LineSeg(ps[0],tb);
//    if(sgn(tmp,0)==0)
//        return getdis(ps[0],ta)<getdis(ps[0],tb);
//    else if(tmp>0)
//        return 1;
return 0;
}
void graham(Point ps[],Point tb[],int n,int &num)
{
int cur=0,top=2;
for(int i=1;i<n;i++)
if(sgn(ps[cur].y,ps[i].y)>0 || (sgn(ps[cur].y,ps[i].y)==0 && sgn(ps[cur].x,ps[i].x)>0))
cur=i;
swap(ps[cur],ps[0]);
sort(ps+1,ps+n,cmp);
tb[0]=ps[0],tb[1]=ps[1],tb[2]=ps[2];
for(int i=3;i<n;i++)
{
while(sgn(LineSeg(tb[top-1],tb[top])/LineSeg(tb[top-1],ps[i]),0)<0)
top--;
tb[++top]=ps[i];
}
num=top+1;
}
/** 卷包裹法求点集凸壳，参数说明同graham算法 **/
void ConvexClosure(Point ps[],Point tb[],int n,int &num)
{
LineSeg lx,ly;
int cur,ch;
bool vis[Max];
num=-1,cur=0;
memset(vis,0,sizeof(vis));
for(int i=1;i<n;i++)
if(sgn(ps[cur].y,ps[i].y)>0 || (sgn(ps[cur].y,ps[i].y)==0 && sgn(ps[cur].x,ps[i].x)>0))
cur=i;
tb[++num]=ps[cur];
lx.s=Point(ps[cur].x-1,ps[cur].y),lx.e=ps[cur];
/// 选取与最后一条确定边夹角最小的点，即余弦值最大者
while(1)
{
double mxcross=-2,midis,tmxcross;
ly.s=lx.e;
for(int i=0;i<n;i++)if(!vis[i])
{
ly.e=ps[i];
tmxcross=(lx*ly)/lx.lenth()/ly.lenth();
if(sgn(tmxcross,mxcross)>0 ||(sgn(tmxcross,mxcross)==0 && getdis(ly.s,ly.e)<midis))
mxcross=tmxcross,midis=getdis(ly.s,ly.e),ch=i;
}
if(ch==cur)break;
tb[++num]=ps[ch],vis[ch]=1;
lx.s=tb[num-1],lx.e=tb[num],ly.s=tb[num];
}
}