poj2826An Easy Problem?!

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繁琐细节题。

1、线段无交点时，ans=0;

2、如图 假设过p3.y的水平线与p1p2相交

因为雨是垂直下落的，左图的情况是无法收集到雨水的

  1 #include <iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<stdlib.h>
  6 #include<vector>
  7 #include<cmath>
  8 #include<queue>
  9 #include<set>
 10 using namespace std;
 11 #define N 100000
 12 #define LL long long
 13 #define INF 0xfffffff
 14 const double eps = 1e-8;
 15 const double pi = acos(-1.0);
 16 const double inf = ~0u>>2;
 17 struct point
 18 {
 19     double x,y;
 20     point(double x=0,double y=0):x(x),y(y) {}
 21 } p[10];
 22 typedef point pointt;
 23 pointt operator -(point a,point b)
 24 {
 25     return pointt(a.x-b.x,a.y-b.y);
 26 }
 27 int dcmp(double x)
 28 {
 29     if(fabs(x)<eps) return 0;
 30     return x<0?-1:1;
 31 }
 32 double cross(point a,point b)
 33 {
 34     return a.x*b.y-a.y*b.x;
 35 }
 36 bool cmp(point a,point b)
 37 {
 38     return a.y>b.y;
 39 }
 40
 41 bool intersection1(point p1, point p2, point p3, point p4, point& p)      // 直线相交
 42 {
 43     double a1, b1, c1, a2, b2, c2, d;
 44     a1 = p1.y - p2.y;
 45     b1 = p2.x - p1.x;
 46     c1 = p1.x*p2.y - p2.x*p1.y;
 47     a2 = p3.y - p4.y;
 48     b2 = p4.x - p3.x;
 49     c2 = p3.x*p4.y - p4.x*p3.y;
 50     d = a1*b2 - a2*b1;
 51     if (!dcmp(d))    return false;
 52     p.x = (-c1*b2 + c2*b1) / d;
 53     p.y = (-a1*c2 + a2*c1) / d;
 54     return true;
 55 }
 56 double solve(point p1,point p2,point p3,point p4,point pp,point tp)
 57 {
 58     double ans;
 59     point p11 = point(p1.x,p1.y+1);
 60     if(dcmp(cross(p11-p1,p1-p3))==0)
 61     {
 62         ans = 0;
 63         //cout<<",";
 64     }
 65     else
 66     {
 67         if(dcmp(cross(p11-p1,p1-p3))*dcmp(cross(p1-p2,p3-p4))<0)
 68         {
 69             ans = fabs(cross(p3-pp,tp-pp))/2;
 70             //cout<<",";
 71         }
 72         else ans = 0;
 73     }
 74     return ans;
 75 }
 76 int on_segment( point p1,point p2 ,point p )
 77 {
 78     double max=p1.x > p2.x ? p1.x : p2.x ;
 79     double min =p1.x < p2.x ? p1.x : p2.x ;
 80     double max1=p1.y > p2.y ? p1.y : p2.y ;
 81     double min1=p1.y < p2.y ? p1.y : p2.y ;
 82     if( p.x >=min && p.x <=max &&
 83             p.y >=min1 && p.y <=max1 )
 84         return 1;
 85     else
 86         return 0;
 87 }
 88 int main()
 89 {
 90     int t,i;
 91     cin>>t;
 92     while(t--)
 93     {
 94         for(i = 1; i <= 4 ; i++)
 95             scanf("%lf%lf",&p[i].x,&p[i].y);
 96         sort(p+1,p+3,cmp);
 97         sort(p+3,p+5,cmp);
 98         point pp;
 99         if(intersection1(p[1],p[2],p[3],p[4],pp))
100         {
101             if(!on_segment(p[1],p[2],pp)||!on_segment(p[3],p[4],pp))
102             {
103                 puts("0.00");
104                 continue;
105             }
106         }
107         //cout<<pp.x<<" "<<pp.y<<endl;
108         double ans;
109         point p1 = point(p[3].x-1,p[3].y),p2;
110         point p3 = point(p[1].x-1,p[1].y);
111         if(intersection1(p[1],p[2],p[3],p1,p2)&&on_segment(p[1],p[2],p2))
112         {
113
114             ans = solve(p[1],p[2],p[3],p[4],pp,p2);
115         }
116         else
117         {
118             intersection1(p[3],p[4],p[1],p3,p2);
119            // cout<<p2.x<<" "<<p2.y<<endl;
120             ans = solve(p[3],p[4],p[1],p[2],pp,p2);
121         }
122         printf("%.2f\n",ans+eps);
123
124     }
125     return 0;
126 }

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