hdu 1348(凸包)

 1 /*
 2 *  凸包
 3 *  注意：给出的点为整点数，不要设某一精度值！ 
 4 */
 5 
 6 #include <cmath>
 7 #include <cstdio>
 8 #include <cstdlib>
 9 #include <iostream>
10 
11 using namespace std;
12 
13 const int N = 1005;
14 const double PI = 3.1415927;
15 
16 struct point {
17     double x;
18     double y;
19 }p[N], stack[N];
20 
21 double dis(point A, point B) {
22     return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));
23 }
24 
25 double crossProd(point A, point B, point C) {
26     return (B.x-A.x)*(C.y-A.y) - (B.y-A.y)*(C.x-A.x);
27 }
28 
29 //以最左下的点为基准点，其他各点（逆时针方向）以极角从小到大的排序规则
30 int cmp(const void *a, const void *b) {
31     point *c = (point *)a;
32     point *d = (point *)b;
33     double k = crossProd(p[0], *c, *d);//极角大小转化为求叉乘
34     if (k<0 || !k && dis(p[0], *c)>dis(p[0], *d)) return 1;
35     return -1;
36 }
37 
38 double Graham(int n) {
39     double x = p[0].x;
40     double y = p[0].y;
41     int mi = 0;
42     for (int i=1; i<n; ++i) {//找到最左下的一个点
43         if (p[i].x<x || (p[i].x==x && p[i].y<y)) {
44             x = p[i].x;
45             y = p[i].y;
46             mi = i;
47         }
48     }
49     point tmp = p[mi];
50     p[mi] = p[0];
51     p[0] = tmp;
52     qsort(p+1, n-1, sizeof(point), cmp);
53     p[n] = p[0];
54     stack[0] = p[0];
55     stack[1] = p[1];
56     stack[2] = p[2];
57     int top = 2;
58     for (int i=3; i<=n; ++i) {//加入一个点后,向右偏拐或共线，则上一个点不在凸包内，则--top，该过程直到不向右偏拐或没有三点共线的点 
59         while (crossProd(stack[top-1], stack[top], p[i])<=0 && top>=2) --top;
60         stack[++top] = p[i];//在当前情况下符合凸包的点，入栈
61     }
62     double len = 0;
63     for (int i=0; i<top; ++i) len += dis(stack[i], stack[i+1]);
64     return len;
65 }
66 
67 int main() {
68     int t;
69     while (scanf("%d", &t) != EOF) {
70         while (t--) {
71             int n, l;
72             scanf ("%d%d", &n, &l);
73             for (int i=0; i<n; ++i) scanf ("%lf%lf", &p[i].x, &p[i].y);
74             double ans = Graham(n);
75             ans += PI * (l + l);
76             printf ("%.0lf\n", ans);
77             if (t) printf ("\n");
78         }
79     }
80     return 0;
81 }