题意：给定一些点，求围住这些点的墙的长度，强要求与这些点至少距离为R
思路：凸包，然后还要加上一个圆周长（画一下就知道了）
code：
1 /* 2 State:Accepted 3 Time:2013-03-24 20:58:46 4 */ 5 #include <iostream> 6 #include <cstdlib> 7 #include <cstdio> 8 #include <cstring> 9 #include <fstream> 10 #include <cmath> 11 #include <algorithm> 12 using namespace std; 13 struct oo{ int x, y; }; 14 int n ,L, xt, yt; 15 oo a[1500], q[2000]; 16 17 void init(){ 18 scanf("%d%d",&n ,&L); 19 for (int i = 1; i <= n ; ++i) 20 scanf("%d%d",&a[i].x , &a[i].y); 21 22 } 23 24 bool cmp(const oo a, const oo b){ 25 if (a.x <b.x || a.x == b.x && a.y < b.y) return true; 26 return false; 27 } 28 29 bool cmp1(const oo a , const oo b ){ 30 int x1 = a.x - xt; 31 int x2 = b.x - xt; 32 int y1 = a.y - yt; 33 int y2 = b.y - yt; 34 if (x1*y2 > x2*y1) return true; 35 if (x1*y2 == x2*y1 && x1*x1 + y1*y1 < x2*x2 + y2*y2) return true; 36 return false; 37 38 } 39 40 void solve(){ 41 sort(a+1, a+n+1,cmp); 42 xt = a[1].x; 43 yt = a[1].y; 44 sort(a+2, a+n+1,cmp1); 45 int h = 1, t = 2; 46 q[h] = a[1]; 47 q[t] = a[2]; 48 int x2 , y2 , x1, y1; 49 for (int i = 3; i <= n ;++i){ 50 while (h < t){ 51 x1 = a[i].x - q[t - 1].x; 52 y1 = a[i].y - q[t - 1].y; 53 x2 = q[t].x - q[t - 1].x; 54 y2 = q[t].y - q[t - 1].y; 55 if (x1*y2 - x2*y1 >= 0) --t; 56 else break; 57 } 58 q[++t] = a[i]; 59 } 60 61 double ans = 0 ,dxy; 62 q[++t] = q[1]; 63 for (int i = 2; i <= t; ++i){ 64 x1 = q[i].x - q[i-1].x; 65 y1 = q[i].y - q[i-1].y; 66 dxy = x1*x1 + y1*y1; 67 ans += pow(dxy, 0.5); 68 } 69 ans += double(3.14159265358979)*L*2; 70 printf("%.0f\n",ans); 71 } 72 73 int main(){ 74 freopen("poj1113.in","r",stdin); 75 freopen("poj1113.out","w",stdout); 76 init(); 77 solve(); 78 fclose(stdin); fclose(stdout); 79 }