POJ 1113&&HDU 1348
题意：凸包周长+一个完整的圆周长。因为走一圈，经过拐点时，所形成的扇形的内角和是360度，故一个完整的圆。
模板题，之前写的Graham模板不对，WR了很多发。。。。POJ上的AC代码
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cstring> 5 #include<set> 6 #include<stdio.h> 7 #include<stdlib.h> 8 #include<math.h> 9 #define clc(a,b) memset(a,b,sizeof(a)) 10 #define eps 1e-8 11 #define PI acos(-1.0) 12 typedef long long LL; 13 const int mod=47000; 14 const int inf=0x3f3f3f3f; 15 const int MAXN=1000; 16 using namespace std; 17 18 const int N = 1005; 19 20 int top; 21 struct point 22 { 23 double x; 24 double y; 25 } p[N], Stack[N]; 26 27 double dis(point A, point B) 28 { 29 return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y)); 30 } 31 32 double crossProd(point A, point B, point C) 33 { 34 return (B.x-A.x)*(C.y-A.y) - (B.y-A.y)*(C.x-A.x); 35 } 36 37 38 int cmp(const void *a, const void *b) 39 { 40 point *c = (point *)a; 41 point *d = (point *)b; 42 double k = crossProd(p[0], *c, *d); 43 if (k<0 || !k && dis(p[0], *c)>dis(p[0], *d)) 44 return 1; 45 return -1; 46 } 47 48 void Graham(int n) 49 { 50 double x = p[0].x; 51 double y = p[0].y; 52 int mi = 0; 53 for (int i=1; i<n; ++i) 54 { 55 if (p[i].x<x || (p[i].x==x && p[i].y<y)) 56 { 57 x = p[i].x; 58 y = p[i].y; 59 mi = i; 60 } 61 } 62 point tmp = p[mi]; 63 p[mi] = p[0]; 64 p[0] = tmp; 65 qsort(p+1, n-1, sizeof(point), cmp); 66 p[n] = p[0]; 67 Stack[0] = p[0]; 68 Stack[1] = p[1]; 69 Stack[2] = p[2]; 70 top = 2; 71 for (int i=3; i<=n; ++i) 72 { 73 while (crossProd(Stack[top-1], Stack[top], p[i])<=0 && top>=2) --top; 74 Stack[++top] = p[i]; 75 } 76 } 77 78 int main() 79 { 80 int n, l; 81 while (scanf("%d%d", &n, &l) != EOF) 82 { 83 for (int i=0; i<n; ++i) 84 { 85 scanf ("%lf%lf", &p[i].x, &p[i].y); 86 } 87 Graham(n); 88 double ans=0; 89 for(int i=0;i<top;i++) 90 { 91 ans+=dis(Stack[i], Stack[i+1]); 92 } 93 ans += PI * (l + l); 94 printf ("%d\n", (int)(ans+0.5)); 95 } 96 return 0; 97 }
