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  1. #include <iostream>
  2. using namespace std;
  3.  
  4. int position(double x ,double y, double windows[]){
  5.     if(x<windows[0]){
  6.         if( y<windows[1] )
  7.             return 5;
  8.         else if( y>=windows[1] && y<=windows[3])
  9.             return 4;
  10.         else if( y>windows[3] )
  11.             return 3;
  12.     }
  13.     else if(x>=windows[0] && x<=windows[2] ){
  14.         if( y<windows[1] )
  15.             return 2;
  16.         else if( y>=windows[1] && y<=windows[3])
  17.             return 0;
  18.         else if( y>windows[3] )
  19.             return 1;
  20.     }
  21.     else if(x>windows[2]){
  22.         if( y<windows[1] )
  23.             return 8;
  24.         else if( y>=windows[1] && y<=windows[3])
  25.             return 7;
  26.         else if( y>windows[3] )
  27.             return 6;
  28.     }
  29.     return -1;
  30. }
  31.  
  32. int main() {
  33.     double xl=0, yd=0, xr=0, yu=0;
  34.     cin>>xl>>yd>>xr>>yu;
  35.     double windows[4] = {xl, yd, xr, yu};
  36.     int m;
  37.     cin>>m;
  38.     for(int i=1;i<=m;i++){
  39.         double x1, y1, x2, y2;
  40.     //第一步，排除性测试
  41.         if( x1<xl && x2<xl )
  42.             { cout<<"null";   break; }
  43.         else if ( x1>xr && x2>xr )
  44.             { cout<<"null";   break; }
  45.         else if ( y1<yd && y2<yd )
  46.             { cout<<"null";   break; }
  47.         else if ( y1>yu && y2>yu )
  48.             { cout<<"null";   break; }
  49.     //第二步，判断线段是否完全在区域内
  50.         int pos1 = position(x1,y1,windows);
  51.         int pos2 = position(x2,y2,windows);
  52.         if (pos1 == 0 && pos2 == 0)
  53.             {cout<<x1<<" "<<y1<<" "<<x2<<" "<<y2; break;}
  54.     //第三步，若A点在xl左侧,将A一侧裁减掉
  55.         if( x1<xl ){
  56.             double y;
  57.             y1 == y2 ? y = y1 : y = y1 + (xl-x1)*(y2-y1)/(x2-x1);
  58.             if ( y>=yd && y<=yu)
  59.                 { x1 = xl; y1 = y;}
  60.             else if (pos1 == 4)
  61.                 { cout<<"null";   break; }
  62.             else if (pos1 == 5){
  63.                 double x = x1 +(yd-y1)*(x2-x1)/(y2-y1);
  64.                 if ( x<=xr && x>=xl)
  65.                     { x1 = x; y1 = yd;}
  66.                 else
  67.                     { cout<<"null";   break; }
  68.             }
  69.             else if (pos1 == 3){
  70.                 double x = x1 +(yu-y1)*(x2-x1)/(y2-y1);
  71.                 if ( x<=xr && x>=xl)
  72.                     { x1 = x; y1 = yu;}
  73.                 else
  74.                     { cout<<"null";   break; }
  75.             }
  76.         }
  77.         pos1 = position(x1,y1,windows);
  78.         pos2 = position(x2,y2,windows);
  79.         if (pos1 == 0 && pos2 == 0)
  80.             {cout<<x1<<" "<<y1<<" "<<x2<<" "<<y2; break;}
  81.     //第四步，若B点在xr右侧,将B一侧裁减掉
  82.         if( x2>xr ){
  83.             double y;
  84.             y1 == y2 ? y = y1 : y = y1 + (xr-x1)*(y2-y1)/(x2-x1);
  85.             if ( y>=yd && y<=yu)
  86.                 { x2 = xr; y2 = y;}
  87.             else if (pos2 == 7)
  88.                 { cout<<"null";   break; }
  89.             else if (pos1 == 8){
  90.                 double x = x1 +(yd-y1)*(x2-x1)/(y2-y1);
  91.                 if ( x<=xr && x>=xl)
  92.                     { x2 = x; y2 = yd;}
  93.                 else
  94.                     { cout<<"null";   break; }
  95.             }
  96.             else if (pos1 == 6){
  97.                 double x = x1 +(yu-y1)*(x2-x1)/(y2-y1);
  98.                 if ( x<=xr && x>=xl)
  99.                     { x2 = x; y2 = yu;}
  100.                 else
  101.                     { cout<<"null";   break; }
  102.             }
  103.         }
  104.         pos1 = position(x1,y1,windows);
  105.         pos2 = position(x2,y2,windows);
  106.         if (pos1 == 0 && pos2 == 0)
  107.             {cout<<x1<<" "<<y1<<" "<<x2<<" "<<y2; break;};
  108.     //第五步，减去两端之后，如果还没有完成,说明两个点都在中间0,1,2,里面：
  109.         if(y1>=yu){
  110.             double x ;
  111.             x1 == x2 ? x = x1 :x = x1 +(yu-y1)*(x2-x1)/(y2-y1);
  112.             x1 = x; y1 = yu;
  113.         };
  114.         if(y1<=yd){
  115.             double x ;
  116.             x1 == x2 ? x = x1 :x = x1 +(yd-y1)*(x2-x1)/(y2-y1);
  117.             x1 = x; y1 = yd;
  118.         };
  119.         if(y2>=yu){
  120.             double x ;
  121.             x1 == x2 ? x = x1 :x = x1 +(yu-y1)*(x2-x1)/(y2-y1);
  122.             x1 = x; y2 = yu;
  123.         };
  124.         if(y2<=yd){
  125.             double x ;
  126.             x1 == x2 ? x = x1 :x = x1 +(yd-y1)*(x2-x1)/(y2-y1);
  127.             x1 = x; y2 = yd;
  128.         };
  129.         pos1 = position(x1,y1,windows);
  130.         pos2 = position(x2,y2,windows);
  131.         if (pos1 == 0 && pos2 == 0)
  132.             {cout<<x1<<" "<<y1<<" "<<x2<<" "<<y2; break;}
  133.         cout<<"Fail";
  134.     }
  135. 	return 0;
  136. }
Success #stdin #stdout 0s 4932KB
comments (?)
stdin 
1 1 5 5
2
0 0 3 4
2 2 3 3
stdout 
null
https://ideone.com/mLINFj
language:
C++ (gcc 8.3)
created:
4 years ago
visibility:
public

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