• 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 }


     

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  • 原文地址:https://www.cnblogs.com/ygdblogs/p/4929884.html
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