poj 2826

太自闭了。。。明明很水的说。。。

有三种为0的情况。一种是有水平，一种是没交点。

第三种有点难想啊。就是下面那个板子被上面的挡住了，，雨水进不去。。。

关键是我翻了翻discuss找到了这种数据，，但是脑子短路了不知道怎么写。。。

我们只要把下面那个端点向上延长！看和上面的板子有没有交点就行了啊！还是看了kuangbin的代码受到的启发。。。

哇我是傻逼啊。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cmath>
 4 #include <iostream>
 5 #include <iomanip>
 6 
 7 using namespace std;
 8 typedef double db;
 9 const db eps = 1e-6;
10 const db pi = acos(-1);
11 int sign(db k){if(k>eps)return 1; else if(k<-eps)return -1;return 0;}
12 int cmp(db k1,db k2){ return sign(k1-k2);}
13 struct point{
14     db x,y;
15     point operator+(const point &k1)const { return point{k1.x+x,k1.y+y};}
16     point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
17     point operator * (db k1) const{return (point){x*k1,y*k1};}
18     point operator / (db k1) const{return (point){x/k1,y/k1};}
19     db abs(){ return sqrt(x*x+y*y);}
20 };
21 db cross(point k1,point k2){ return k1.x*k2.y-k1.y*k2.x;}
22 db dot(point k1,point k2){ return k1.x*k2.x+k1.y*k2.y;}
23 int intersect(db l1,db r1,db l2,db r2){
24     if (l1>r1) swap(l1,r1); if (l2>r2) swap(l2,r2); return cmp(r1,l2)!=-1&&cmp(r2,l1)!=-1;
25 }
26 int checkSS(point k1,point k2,point k3,point k4){
27     return intersect(k1.x,k2.x,k3.x,k4.x)&&intersect(k1.y,k2.y,k3.y,k4.y)&&
28            sign(cross(k3-k1,k4-k1))*sign(cross(k3-k2,k4-k2))<=0&&
29            sign(cross(k1-k3,k2-k3))*sign(cross(k1-k4,k2-k4))<=0;
30 }
31 point getLL(point k1,point k2,point k3,point k4){
32     db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3); return (k1*w2+k2*w1)/(w1+w2);
33 }
34 point p[6];
35 int t;
36 int main(){
37     ios::sync_with_stdio(false);
38     cout<<fixed<<setprecision(2);
39     cin>>t;
40     while (t--) {
41         for (int i = 1; i <= 4; i++) {
42             cin >> p[i].x >> p[i].y;
43         }
44         bool f=1;
45         if(cmp(p[1].y,p[2].y)==0)f=0;
46         if(cmp(p[3].y,p[4].y)==0)f=0;
47         if(!checkSS(p[1],p[2],p[3],p[4]))f=0;
48         if(!f){
49             cout<<0.00<<endl;
50             continue;
51         }
52         point tmp = getLL(p[1],p[2],p[3],p[4]);
53         if(p[1].y<p[2].y) swap(p[1],p[2]);
54         if(p[3].y<p[4].y) swap(p[3],p[4]);
55         if(p[1].y>p[3].y){//maxy=p[1]
56             point ano = getLL(p[1],tmp,p[3],{p[3].x-1,p[3].y});
57             if (checkSS({p[3].x,p[3].y},{p[3].x,20000},ano,p[1])){
58                 cout<<0.00<<endl;
59                 continue;
60             }
61             db ans = cross(p[3]-tmp,ano-tmp)/2;
62             ans=abs(ans);
63             cout<<ans<<endl;
64         } else{
65             point ano = getLL(p[3],tmp,p[1],{p[1].x-1,p[1].y});
66             if (checkSS({p[1].x,p[1].y},{p[1].x,20000},ano,p[3])){
67                 cout<<0.00<<endl;
68                 continue;
69             }
70             db ans = cross(p[1]-tmp,ano-tmp)/2;
71             ans=abs(ans);
72             cout<<ans<<endl;
73         }
74     }
75 }
76 /**
77 1
78 0 0 10 10
79 0 0 8 9
80  */

View Code

 

标签：point,k3,k2,k1,poj,k4,2826,include
来源： https://www.cnblogs.com/MXang/p/10439934.html