Poj 1113 Wall [凸包]

题目大意：

给定多边形城堡的n个顶点，绕城堡外面建一个围墙，围住所有点，并且墙与所有点的距离至少为L，求这个墙最小的长度

思路：

结果等于凸包周长 + 一个完整圆的周长。

当绕城堡外面走一圈，在拐角处所形成的弧度相加为一个长度为Ｌ的圆。

  1 /*
  2     poj 1113
  3     凸包 
  4     大意：
  5     给定多边形城堡的n个顶点，绕城堡外面建一个围墙，围住所有点，
  6     并且墙与所有点的距离至少为L，求这个墙最小的长度 
  7 */
  8 
  9 #include <iostream>
 10 #include <math.h>
 11 #include <cstring> 
 12 #include <algorithm>
 13 #include <stdlib.h>
 14 
 15 using namespace std;
 16 #define pi acos(-1.0)
 17 #define eps 1e-8
 18 #define zero(x) (((x)>0?(x):-(x))<eps)
 19 
 20 struct point
 21 {
 22     double x, y;
 23 }p[1005], convex[1005];
 24 
 25 //两点间距离 
 26 double dis(point a, point b)
 27 {
 28     return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y) * (a.y - b.y));
 29 }
 30 
 31 //计算cross product (P1-P0)x(P2-P0)
 32 //叉积 
 33 double xmult(point p1, point p2, point p0)
 34 {
 35     return (p1.x - p0.x)*(p2.y - p0.y) - (p2.x - p0.x)*(p1.y - p0.y);
 36 }
 37 //graham算法顺时针构造包含所有共线点的凸包,O(nlogn)
 38 point p1, p2;
 39 
 40 int graham_cp(const void* a, const void* b)
 41 {
 42     double ret = xmult(*((point*) a), *((point*) b), p1);
 43     double ret2 =  xmult(*((point*) a), *((point*) b), p2);
 44     return zero(ret) ? (ret2> 0 ? 1 : -1) : (ret > 0 ? 1 : -1);
 45 }
 46 
 47 void _graham(int n, point* p, int& s, point* ch)
 48 {
 49     int i, k = 0;
 50     for (p1 = p2 = p[0], i = 1; i<n; p2.x += p[i].x, p2.y += p[i].y, i++)
 51         if (p1.y - p[i].y>eps || (zero(p1.y - p[i].y) && p1.x > p[i].x))
 52             p1 = p[k = i];
 53     p2.x /= n, p2.y /= n;
 54     p[k] = p[0], p[0] = p1;
 55     qsort(p + 1, n - 1, sizeof(point), graham_cp);
 56     for (ch[0] = p[0], ch[1] = p[1], ch[2] = p[2], s = i = 3; i < n; ch[s++] = p[i++])
 57         for (; s>2 && xmult(ch[s - 2], p[i], ch[s - 1]) < -eps; s--);
 58 }
 59 
 60 int wipesame_cp(const void *a, const void *b)
 61 {
 62     if ((*(point *) a).y < (*(point *) b).y - eps) return -1;
 63     else if ((*(point *) a).y > (*(point *) b).y + eps) return 1;
 64     else if ((*(point *) a).x < (*(point *) b).x - eps) return -1;
 65     else if ((*(point *) a).x > (*(point *) b).x + eps) return 1;
 66     else return 0;
 67 }
 68 
 69 int _wipesame(point * p, int n)
 70 {
 71     int i, k;
 72     qsort(p, n, sizeof(point), wipesame_cp);
 73     for (k = i = 1; i < n; i++)
 74         if (wipesame_cp(p + i, p + i - 1) != 0) p[k++] = p[i];
 75     return k;
 76 }
 77 
 78 //构造凸包接口函数,传入原始点集大小n,点集p(p原有顺序被打乱!)
 79 //返回凸包大小,凸包的点在convex中
 80 //参数maxsize为1包含共线点,为0不包含共线点,缺省为1
 81 //参数dir为1顺时针构造,为0逆时针构造,缺省为1
 82 //在输入仅有若干共线点时算法不稳定,可能有此类情况请另行处理!
 83 int graham(int n, point* p, point* convex, int maxsize = 1, int dir = 1)
 84 {
 85     point* temp = new point[n];
 86     int s, i;
 87     n = _wipesame(p, n);
 88     _graham(n, p, s, temp);
 89     for (convex[0] = temp[0], n = 1, i = (dir ? 1 : (s - 1)); dir ? (i < s) : i; i += (dir ? 1 : -1))
 90         if (maxsize || !zero(xmult(temp[i - 1], temp[i], temp[(i + 1)%s])))
 91             convex[n++] = temp[i];
 92     delete []temp;
 93     return n;
 94 }
 95 
 96 int main()
 97 {
 98     int n;
 99     double d;
100     while (cin >> n >> d)
101     {
102         memset(p,0,sizeof(p));
103         memset(convex, 0, sizeof(convex));
104         double dist = 0.0;
105         for (int i = 0; i < n; i++)
106         {
107             cin >> p[i].x >> p[i].y;
108         }
109         //cout << n << endl;
110         int size = graham(n, p, convex,0);
111 //        cout << size << endl;
112         for (int i = 0; i < size - 1; i++)
113         {
114             dist += dis(convex[i], convex[i + 1]);
115         }
116         dist += dis(convex[size - 1], convex[0]);//别忘了起始到结束的长度
117         cout << (int) (dist + 2 * pi * d + 0.5) << endl;
118     }
119 }