Wall--POJ 1113

1、题目类型：凸包。

2、解题思路：（1）建立点集map[][]；（2）Graham法求解凸包，中间对于斜率的排序用的是STL的sort()；（3）求解凸包边长的和加上以L为周长的园的周长。

3、注意事项：凸包处理过程中，Graham(),用于记录的结构体数组的开始与结束。

4、实现方法:

  
    

   
     
   
     #include
   
     <
   
     iostream
   
     >
   
     
#include
   
     <
   
     algorithm
   
     >
   
     
#include
   
     <
   
     cmath
   
     >
   
     

   
     #define
   
      PI 3.1415926
   
     

   
     using
   
      
   
     namespace
   
      std;


   
     struct
   
      Point
{
 
   
     int
   
      x,y;
};


   
     int
   
      n,L;
Point map[
   
     1010
   
     ];
Point S[
   
     1010
   
     ];


   
     //
   
     求斜率
   
     

   
     float
   
      XieLv(Point p1,Point p2,Point p0)
{
 
   
     return
   
      ((p1.x
   
     -
   
     p0.x)
   
     *
   
     (p2.y
   
     -
   
     p0.y)
   
     -
   
     (p2.x
   
     -
   
     p0.x)
   
     *
   
     (p1.y
   
     -
   
     p0.y));
}


   
     //
   
     两点距离
   
     

   
     double
   
      Dis(Point p1,Point p2)
{
 
   
     double
   
      distance
   
     =
   
     sqrt((
   
     double
   
     )(p1.x
   
     -
   
     p2.x)
   
     *
   
     (p1.x
   
     -
   
     p2.x)
   
     +
   
     (p1.y
   
     -
   
     p2.y)
   
     *
   
     (p1.y
   
     -
   
     p2.y));
 
   
     return
   
      distance;
}


   
     int
   
      cmp(Point p1,Point p2)
{
 
   
     if
   
     (XieLv(p1,p2,map[
   
     0
   
     ])
   
     >
   
     0
   
     )
 
   
     return
   
      
   
     1
   
     ;
 
   
     if
   
     ((XieLv(p1,p2,map[
   
     0
   
     ])
   
     ==
   
     0
   
     &&
   
     
 Dis(p1,map[
   
     0
   
     ])
   
     -
   
     Dis(p2,map[
   
     0
   
     ])
   
     <
   
     0
   
     ))
 
   
     return
   
      
   
     1
   
     ;
 
   
     return
   
      
   
     0
   
     ;
}


   
     //
   
     Graham法求凸包，返回凸包中点数
   
     

   
     int
   
      Graham()
{
 
   
     int
   
      i,top,tmp
   
     =
   
     0
   
     ;
 
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )
 {
 
   
     if
   
     ((map[i].y
   
     <
   
     map[tmp].y)
   
     ||
   
     (map[i].y
   
     ==
   
     map[tmp].y
   
     &&
   
     map[i].x
   
     <
   
     map[tmp].x))
 tmp
   
     =
   
     i;
 }
 Point p
   
     =
   
     map[tmp];
 map[tmp]
   
     =
   
     map[
   
     0
   
     ];
 map[
   
     0
   
     ]
   
     =
   
     p;
 sort(map
   
     +
   
     1
   
     ,map
   
     +
   
     n,cmp); 
 top
   
     =
   
     2
   
     ;
 S[
   
     0
   
     ]
   
     =
   
     map[
   
     0
   
     ];
 S[
   
     1
   
     ]
   
     =
   
     map[
   
     1
   
     ];
 S[
   
     2
   
     ]
   
     =
   
     map[
   
     2
   
     ];
 
   
     for
   
     (i
   
     =
   
     3
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )
 {
 
   
     while
   
     (XieLv(map[i],S[top],S[top
   
     -
   
     1
   
     ])
   
     >
   
     0
   
     )
 top
   
     --
   
     ;
 S[
   
     ++
   
     top]
   
     =
   
     map[i];
 }
 
   
     return
   
      top
   
     +
   
     1
   
     ;
}


   
     int
   
      main()
{
 
   
     int
   
      i,num;
 
   
     double
   
      ans
   
     =
   
     0.0
   
     ;
 cin
   
     >>
   
     n
   
     >>
   
     L;
 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )
 {
 scanf(
   
     "
   
     %d %d
   
     "
   
     ,
   
     &
   
     map[i].x,
   
     &
   
     map[i].y);
 }
 num
   
     =
   
     Graham();
 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     num
   
     -
   
     1
   
     ;i
   
     ++
   
     )
 {
 ans
   
     +=
   
     Dis(S[i],S[i
   
     +
   
     1
   
     ]);
 }
 ans
   
     +=
   
     Dis(S[num
   
     -
   
     1
   
     ],S[
   
     0
   
     ]);
 ans
   
     =
   
     ans
   
     +
   
     2
   
     *
   
     L
   
     *
   
     PI;
 printf(
   
     "
   
     %.0f\n
   
     "
   
     ,ans);
 
   
     return
   
      
   
     0
   
     ;
}