题目链接：https://vjudge.net/problem/POJ-2074
坑点：障碍物得在up直线和down直线之间。ve得多加一个down的终点，因为可能全部都能看到。最后用float，不要用double。。。。。我也不知道为什么。。。
1 /************************************************************************* 2 > File Name: poj2074.cpp 3 # File Name: poj2074.cpp 4 # Author : xiaobuxie 5 # QQ : 760427180 6 # Email:[email protected] 7 # Created Time: 2019年09月22日 星期日 16时06分36秒 8 ************************************************************************/ 9 10 #include<iostream> 11 #include<cstdio> 12 #include<map> 13 #include<cmath> 14 #include<cstring> 15 #include<set> 16 #include<queue> 17 #include<vector> 18 #include<algorithm> 19 using namespace std; 20 typedef long long ll; 21 #define inf 0x3f3f3f3f 22 #define pq priority_queue<int,vector<int>,greater<int> > 23 ll gcd(ll a,ll b){ 24 if(a<b) return gcd(b,a); 25 return b==0?a:gcd(b,a%b); 26 } 27 28 #define eps 1e-10 29 #define pii pair < double,double > 30 const int N=1e5+9; 31 int sgn(double x){ 32 if(fabs(x)<eps) return 0; 33 if(x<0) return -1; 34 return 1; 35 } 36 struct Point{ 37 double x,y; 38 Point operator - (const Point& b)const{ 39 return (Point){x-b.x,y-b.y}; 40 } 41 double operator * (const Point& b)const{ 42 return x*b.x+y*b.y; 43 } 44 double operator ^ (const Point& b )const{ 45 return x*b.y-b.x*y; 46 } 47 }; 48 struct Line{ 49 Point s,e; 50 }L[N],up,down; 51 double solve(Line l1,Line l2){ 52 Point res = l1.s; 53 double t = ( (l1.s-l2.s) ^ (l2.s-l2.e))/((l1.s-l1.e)^(l2.s-l2.e)); 54 res.x+=(l1.e.x-l1.s.x)*t; 55 return res.x; 56 } 57 int main(){ 58 double x1,x2,y; 59 int n; 60 while(~scanf("%lf %lf %lf",&x1,&x2,&y)){ 61 if(sgn(x1)==0 && sgn(x2)==0 && sgn(y)==0) break; 62 vector<pii> ve; 63 up=(Line){(Point){x1,y},(Point){x2,y}}; 64 scanf("%lf %lf %lf",&x1,&x2,&y); 65 down=(Line){(Point){x1,y},(Point){x2,y}}; 66 scanf("%d",&n); 67 for(int i=1;i<=n;++i){ 68 scanf("%lf %lf %lf",&x1,&x2,&y); 69 if(sgn(y-up.s.y)>=0 || sgn(down.s.y-y)>=0) continue; 70 double u=solve((Line){up.e,(Point){x1,y}},down); 71 double v=solve((Line){up.s,(Point){x2,y}},down); 72 if(sgn(u-v)>=0) continue; 73 ve.push_back(make_pair(u,v)); 74 } 75 ve.push_back(make_pair(down.e.x,down.e.x)); 76 sort(ve.begin(),ve.end()); 77 double ans=0,pos=down.s.x; 78 for(int i=0;i<ve.size();++i){ 79 x1=ve[i].first,x2=ve[i].second; 80 if(x1-pos>ans) ans=x1-pos; 81 if(sgn(x2-pos)>0){ 82 pos=x2; 83 if(sgn(pos-down.e.x)>0) break; 84 } 85 } 86 if(sgn(ans)==0) puts("No View"); 87 else printf("%.2f\n",ans); 88 } 89 return 0; 90 }