凸包模板 水平序的Graham-Scan
该算法的功能是：
对n个无序的点p[]求凸包，结果在s[0~top-1]内。
作为O(nlgn)的算法，已然到达其下界，实现也不复杂。
分析详见黑书。
模板题： http://poj.org/problem?id=1113
1 #include<map> 2 #include<set> 3 #include<list> 4 #include<cmath> 5 #include<ctime> 6 #include<queue> 7 #include<stack> 8 #include<cctype> 9 #include<cstdio> 10 #include<string> 11 #include<vector> 12 #include<cstdlib> 13 #include<cstring> 14 #include<iostream> 15 #include<algorithm> 16 #define MAXN 1005 17 #define INF 0x3f3f3f3f 18 #define LL long long 19 #define DBL double 20 #define eps 1e-6 21 #define PI acos(-1.0) 22 #define Test() cout<<"Test"<<endl; 23 #define Debug(a) cout<<#a<<" = "<<a<<endl; 24 #define Debug2(a,b) cout<<#a<<" = "<<a<<" , "<<#b<<" = "<<b<<endl; 25 using namespace std; 26 27 struct P{ 28 DBL x, y; 29 }; 30 int n, L; 31 P p[MAXN]; 32 33 bool cmp(P a, P b){ 34 return a.y<b.y || (a.y==b.y && a.x<b.x); 35 } 36 DBL mul(DBL t){ 37 return t*t; 38 } 39 DBL dis(P a, P b){ 40 return sqrt(mul(a.x-b.x)+mul(a.y-b.y)); 41 } 42 int dblcmp(DBL x){ 43 return fabs(x)<eps? 0: x>0? 1: -1; 44 } 45 DBL cross(P a, P b, P c){ //ab X ac 46 return (b.x-a.x)*(c.y-a.y) - (c.x-a.x)*(b.y-a.y); 47 } 48 DBL grahamScan(){ //水平序 49 sort(p, p+n, cmp); 50 int top=-1, s[MAXN], tmp; 51 s[++top]=0, s[++top]=1; 52 for(int i=2; i<n; i++){ // 做右链 53 while(top>0 && dblcmp(cross(p[s[top-1]], p[s[top]], p[i])) <= 0) top --; 54 s[++top] = i; 55 } 56 tmp = top; // 此时的栈顶元素一定是第n个点 57 s[++top]=n-2; 58 for(int i=n-3; i>=0; i--){ // 做左链 59 while(top>tmp && dblcmp(cross(p[s[top-1]], p[s[top]], p[i])) <= 0) top --; 60 s[++top] = i; 61 } // 此时的栈顶元素一定是第1个点,即s[top]=s[0] 62 // s[0~top-1]即为所求凸包 63 DBL res=0; 64 for(int i=0; i<top; i++) 65 res += dis(p[s[i]], p[s[i+1]]); 66 return res+2*PI*L; 67 } 68 69 int main(){ 70 while(cin >> n >> L){ 71 for(int i=0; i<n; i++) 72 scanf("%lf%lf", &p[i].x, &p[i].y); 73 printf("%d\n", (int)(grahamScan()+0.5)); //+0.5 74 } 75 return 0; 76 }
