poj2074Line of Sight(直线相交）

链接

几何细节题。

对于每一个障碍物可以求出它在地产线上的覆盖区间，如下图。

紫色部分即为每个障碍物所覆盖掉的区间，求出所有的，扫描一遍即可。

几个需要注意的地方：直线可能与地产线没有交点，可视区间可能包含地产线的端点，扫描的时候保留当前扫到的最大值。

代码中的数据很经典，供参考。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 10100
 #define LL long long
 #define INF 0xfffffff
 #define zero(x) (((x)>0?(x):-(x))<eps)
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 struct point
 {
     double x,y;
     point(double x=,double y=):x(x),y(y) {}
 } p[N];
 struct line
 {
     point u,v;
 } li[N];
 typedef point pointt;
 pointt operator -(point a,point b)
 {
     return point(a.x-b.x,a.y-b.y);
 }
 int dcmp(double x)
 {
     if(fabs(x)<eps) return ;
     return x<?-:;
 }
 bool cmp(point a,point b)
 {
     if(dcmp(a.x-b.x)==)
         return a.y<b.y;
     return a.x<b.x;
 }

 bool intersection1(point p1, point p2, point p3, point p4, point& p)      // 直线相交
 {
     double a1, b1, c1, a2, b2, c2, d;
     a1 = p1.y - p2.y;
     b1 = p2.x - p1.x;
     c1 = p1.x*p2.y - p2.x*p1.y;
     a2 = p3.y - p4.y;
     b2 = p4.x - p3.x;
     c2 = p3.x*p4.y - p4.x*p3.y;
     d = a1*b2 - a2*b1;
     if (!dcmp(d))    return false;
     p.x = (-c1*b2 + c2*b1) / d;
     p.y = (-a1*c2 + a2*c1) / d;
     return true;
 }
 int main()
 {
     int i,n;
     int x1,x2,y;
     while(scanf("%d%d%d",&x1,&x2,&y)&&x1&&x2&&y)
     {
         li[].u = point(x1,y);
         li[].v = point(x2,y);
         li[].v.y = li[].u.y;
         scanf("%lf%lf%lf",&li[].u.x,&li[].v.x,&li[].u.y);
         li[].v.y = li[].u.y;
         scanf("%d",&n);
         for(i = ; i <= n+; i++)
         {
             scanf("%lf%lf%lf",&li[i].u.x,&li[i].v.x,&li[i].u.y);
             li[i].v.y = li[i].u.y;
         }
         n+=;
         int g = ;
         for(i =  ; i <= n; i++)
         {
             if(dcmp(li[i].u.y-li[].u.y)>=||dcmp(li[i].u.y-li[].u.y)<=) continue;
             point pp;
             intersection1(li[].u,li[i].v,li[].u,li[].v,pp);
             p[++g].y = min(li[].v.x,max(pp.x,li[].u.x));
             intersection1(li[].v,li[i].u,li[].u,li[].v,pp);
             p[g].x = max(li[].u.x,min(li[].v.x,pp.x));
         }
         sort(p+,p+g+,cmp);
         double maxz,ty ;
        // cout<<p[1].x<<" "<<p[1].y<<endl;
         if(g==)
         maxz = li[].v.x-li[].u.x;
         else
         {
             maxz = max(p[].x-li[].u.x,li[].v.x-p[g].y);
             ty = p[].y;
         }

         for(i = ; i < g; i++)
         {
            // printf("%.3f %.3f\n",p[i].x,p[i].y);
             ty = max(ty,p[i].y);
             if(p[i+].x>ty)
             {
                 maxz = max(p[i+].x-ty,maxz);//printf("%.3f %.3f\n",ty,maxz);
                 ty = p[i+].y;
             }

         }
         if(dcmp(maxz)<=)
             puts("No View");
         else
             printf("%.2f\n",maxz);
     }
     return ;
 }
 /*
 2 6 6
 0 15 0
 3
 1 2 1
 3 4 1
 12 13 1
 1 5 5
 0 10 0
 1
 0 15 1
 2 6 6
 0 15 0
 3
 1 2 1
 3 4 1
 12 13 1
 2 6 6
 0 15 0
 4
 1 2 1
 3 4 1
 12 13 1
 1 5 2
 2 6 6
 0 15 0
 2
 0 5 3
 6 15 3
 2 6 6
 0 15 0
 2
 6 10 1
 0 2 1
 2 6 6
 0 15 0
 1
 2 6 7
 2 6 6
 0 15 0
 1
 2 6 7
 2 6 6
 0 15 0
 1
 4 4.5 5.5
 2 6 6
 0 15 0
 16
 0 1 3
 1.5 2 3
 2.5 3 3
 3.5 4 3
 4.5 5 3
 5.5 6 3
 6.5 7 3
 7.5 8 3
 8.5 9 3
 9.5 10 3
 10.5 11 3
 11.5 12 3
 12.5 13 3
 13.5 14 3
 14.5 15 3
 15.5 16 3
 2 6 6
 0 15 0
 16
 0 1 .1
 1.5 2 .1
 2.5 3 .1
 3.5 4 .1
 4.5 5 .1
 5.5 6 .1
 6.5 7 .1
 7.5 8 .1
 8.5 9 .1
 9.5 10 .1
 10.5 11 .1
 11.5 12 .1
 12.5 13 .1
 13.5 14 .1
 14.5 15 .1
 15.5 16 .1
 2 6 6
 0 15 0
 14
 0 1 3
 1.5 2 3
 2.5 3 3
 3.5 4 3
 4.5 5 3
 5.5 6 3
 8.5 9 3
 9.5 10 3
 10.5 11 3
 11.5 12 3
 12.5 13 3
 13.5 14 3
 14.5 15 3
 15.5 16 3
 2 6 6
 0 4000000000 0
 2
 1 2 1
 15 16 3
 2 6 6
 0 15 1
 5
 1 1.5 6
 17 18 1
 3 5 3
 0 20 10
 0 20 0.5

 答案：
 8.80
 No View
 8.80
 6.70
 No View
 4.00
 15.00
 15.00
 No View
 No View
 0.44
 1.00
 3999999970.00
 8.00
 */