poj 2074 Line of Sight (计算几何，细节题)

Line of Sight

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 3935		Accepted: 1229

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:

/*************************************************************************
> File Name: code/poj/2074.cpp
> Author: 111qqz
> Email: rkz2013@126.com
> Created Time: 2015年11月08日 星期日 10时26分28秒
************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#include<cctype>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int N=1E4+6;
int n;

int dblcmp(double d)
{
return d<-eps ?-1:d>eps;
}
struct node
{
double x1,x2,y;
}obs
,house,li;

struct point
{
double x,y;
};

bool cmp (node a,node b )
{
return a.x2<b.x2;
}

int main()
{
#ifndef  ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif

while (scanf("%lf %lf %lf",&house.x1,&house.x2,&house.y)!=EOF)
{
if (dblcmp(house.x1)==0&&dblcmp(house.x2)==0&&dblcmp(house.y)==0) break;
scanf("%lf %lf %lf",&li.x1,&li.x2,&li.y);
scanf("%d",&n);
for ( int i = 0 ; i < n ; i++ ) scanf("%lf %lf %lf",&obs[i].x1,&obs[i].x2,&obs[i].y);

sort(obs,obs+n,cmp);
double sum = -1 ;

point p1,p2;
double k;
bool beblocked = false;
for ( int i = 0 ;i <n-1 ; i++)   //最后一个只算右端点就行
{
if (dblcmp(obs[i].y-house.y)>=0||dblcmp(obs[i].y-li.y)<=0) continue;
else beblocked = true;
if (dblcmp(obs[i].x2-obs[i+1].x1)<0)
{
if (dblcmp(obs[i].x2-house.x1)==0)
{
p1.x=house.x1;
}
else
{
k = (house.y-obs[i].y)*1.0/(house.x1-obs[i].x2);
p1.x = house.x1-(house.y-li.y)/k;

}
if (dblcmp(obs[i+1].x1-house.x2)==0)
{
p2.x=house.x2;
}
else
{
k = (house.y-obs[i+1].y)*1.0/(house.x2-obs[i+1].x1);
p2.x = house.x2-(house.y-li.y)/k;
}
p1.x = max(p1.x,li.x1);  //边界维护
p2.x = min(p2.x,li.x2);
//        cout<<"p1.x:"<<p1.x<<" p2.x"<<p2.x<<endl;
if (dblcmp(p2.x-p1.x)>0)  //求连续的最大段。。。不是求和。。认真读题。
{
sum = max(sum,p2.x-p1.x);
beblocked  = true;
//            cout<<"sum:"<<sum<<endl;
}
}
}
//最后一个特殊处理：
if (dblcmp(house.x1-obs[n-1].x2)==0)
{
p1.x = house.x1;
}
else
{
k = (house.y-obs[n-1].y)/(house.x1-obs[n-1].x2);
p1.x = house.x1-house.y/k;
//        cout<<"k:"<<k<<endl;
}
//    cout<<"p1.x:"<<p1.x<<" li.x2:"<<li.x2<<" minus:"<<li.x2-p1.x<<endl;
if (dblcmp(house.y-obs[n-1].y)>0&&dblcmp(li.y-obs[n-1].y)<0)
sum = max(sum,li.x2-p1.x),beblocked = true;

//第一个要要单独处理

if (dblcmp(house.x2-obs[0].x1)==0)
{
p1.x=house.x2;
}
else
{
k = (house.y-obs[0].y)/(house.x2-obs[0].x1);
p1.x = house.x2-house.y/k;
//        cout<<"k:"<<k<<endl;
}
if (dblcmp(house.y-obs[0].y)>0&&dblcmp(li.y-obs[0].y)<0)
sum = max(sum,p1.x-li.x1),beblocked = true;

if (!beblocked)
{
sum = li.x2-li.x1;
}
if (dblcmp(sum)>0)
{
printf("%.2f\n",sum);
}
else
{
puts("No View");
}

}
#ifndef ONLINE_JUDGE
#endif
fclose(stdin);
return 0;
}

View Code