poj 2074 Line of Sight

//主要思想，遍历每一个障碍，求出它在property line上挡住的范围cover[i],然后线性扫描这些cover，找出最大值。

//此代码一开始g++交不过，c++才能过，后来才知道G++标准的浮点型输出用%f 而不是%lf。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const double EP=1e-8;
struct Point{double x, y;}p;
struct Line{double a, b, c;}pl;
struct Segline{Point s, e;}sg,cover[1002],house,pline;
Line makeline(Point p1, Point p2){
Line t1;
int sign=1;
t1.a=p2.y-p1.y;
if(t1.a<0){
t1.a*=-1;
sign=-1;
}
t1.b=sign*(p1.x-p2.x);
t1.c=sign*(p1.y*p2.x-p1.x*p2.y);
return t1;
}
void lineintersect(Line l1, Line l2, Point &p){
double t1=l1.a*l2.b-l2.a*l1.b;
p.x=(l2.c*l1.b-l1.c*l2.b)/t1;
p.y=(l2.a*l1.c-l1.a*l2.c)/t1;
}
bool cmp(Segline s1, Segline s2){
if(s2.s.x-s1.s.x>-EP)
return true;
return false;
}
int main(){
//freopen("1.txt", "r", stdin);
int i, n, sum;
double len, covery;
while(scanf("%lf %lf %lf",&house.s.x, &house.e.x, &house.s.y)!=EOF){
if(!house.s.x&&!house.e.x&&!house.s.y)
break;
house.e.y=house.s.y;sum=0;
scanf("%lf %lf %lf %d",&pline.s.x, &pline.e.x, &pline.s.y, &n);
pline.e.y=pline.s.y;
pl=makeline(pline.s,pline.e);
for(i=0; i<n; i++){
scanf("%lf %lf %lf", &sg.s.x, &sg.e.x, &sg.s.y);
sg.e.y=sg.s.y;
if(sg.s.y-house.s.y>-EP||pline.s.y-sg.s.y>-EP)
continue;
Line l;
l=makeline(house.e,sg.s);
lineintersect(l, pl, p);
if(p.x>pline.e.x)
continue;
cover[sum].s=p;
l=makeline(house.s,sg.e);
lineintersect(l, pl, p);
if(p.x<pline.s.x)
continue;
cover[sum].e=p;
sum++;
}
if(sum==0){
printf("%.2lf/n",pline.e.x-pline.s.x);
continue;
}
sort(cover,cover+sum,cmp);
if(cover[0].s.x-pline.s.x<EP)
len=0;
else len=cover[0].s.x-pline.s.x;
covery=cover[0].e.x;
for(i=1; i<sum; i++){
if(cover[i].s.x-covery>len)
len=cover[i].s.x-covery;
if(cover[i].e.x>covery)
covery=cover[i].e.x;
}
if(pline.e.x-covery>len)
len=pline.e.x-covery;
if(len<EP)
printf("No View/n");
else printf("%.2f/n", len);
}
return 0;
}