>  已排序
>  cc[i](0<=i<=c-1)为第i个可选点
>  p为前半弧指针，q为后半弧指针，mid为弧的中间点
>  getarea计算三点组成的面积
>  getmax得到两数的较大值
>  area保存着最优值
> 
> 		p = 1;
> 		q = 2;
> 		area = 0;
> 		for (i = 2; i < c - 1; i ++)
> 		{
> 			mid = cc[i] / 2;
> 			while ((p < i - 1) && (cc[p] < mid)) p ++;
> 			now = getmax(getarea(0, cc[i], cc[p]), getarea(0, cc[i], cc[p - 1]));
> 			mid = (n + cc[i]) / 2;
> 			while ((q < c - 1) && (cc[q] < mid)) q ++;
> 			now += getmax(getarea(0, cc[i], cc[q]), getarea(0, cc[i], cc[q - 1]));
> 			area = getmax(area, now);
> 		}

• 分过程 ------> (635) Bluebird 2005-07-21 21:00:33
  • 默认0点坐标为(d,0) (2) Bluebird 2005-07-21 21:03:22
  • wrong! (691) frkstyc 2005-07-21 21:03:36
    • if you find out the problem yourself perhaps you will kill yourself (0) frkstyc 2005-07-21 21:04:41
    • where ............ (0) Bluebird 2005-07-21 21:04:41
      • d>>=1,what if d is odd?????????????????????????????????????????????????? (0) frkstyc 2005-07-21 21:05:06
        • 我靠..................................................... (0) Bluebird 2005-07-21 21:05:45
          • 让我撞死吧，豆腐！！！！！！ (0) Bluebird 2005-07-21 21:07:12
            • ft.. (0) frkstyc 2005-07-21 21:07:50
              • sign........怎么还wa......不会是出现精度问题了吧........ (0) Bluebird 2005-07-21 21:22:18
• 用ABsinC/2求面积 ----> (399) Bluebird 2005-07-21 21:02:19
  • d = d * d / 4.0 .... (2) Bluebird 2005-07-21 21:25:04
  • 我用的是向量差积算面积 (439) windbells 2005-07-21 22:27:07
    • 汗....那不就是行列式.... (0) Bluebird 2005-07-22 12:30:40