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POJ 1113 凸包模板题
上模板。
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>#include <vector>#include <utility>#include <stack>#include <queue>#include <map>#include <deque>#define max(x,y) ((x)>(y)?(x):(y))#define min(x,y) ((x)<(y)?(x):(y))#define INF 0x3f3f3f3f#define MAXN 1005using namespace std;const double eps = 1e-8;const double PI = acos(-1.0);int sgn(double x){ if(fabs(x) < eps) return 0; if(x < 0) return -1; return 1;}struct Point{ double x,y; Point(){} Point(double _x, double _y): x(_x),y(_y) {} Point operator -(const Point &B) const { return Point(x-B.x, y-B.y); } Point operator +(const Point &B) const //向量相加 { return Point(x+B.x, y+B.y); } double operator ^(const Point &B) const //叉积 { return x*B.y - y*B.x; } double operator *(const Point &B) const //点积 { return x*B.x + y*B.y; } bool operator ==(const Point &B) const { return fabs(B.x-x)<eps && fabs(B.y-y)<eps; } bool operator !=(const Point &B) const { return !((*this) == B); } void transXY(double B) //绕原点逆时针旋转B弧度 { double tx = x, ty = y; x = tx*cos(B) - ty*sin(B); y = tx*sin(B) + ty*cos(B); } void input() //读入只能用double读入 { scanf("%lf%lf",&x,&y); }};double dist(Point a, Point b){ return sqrt((a-b)*(a-b));}//求凸包,Graham算法//点的编号0~n-1//返回凸包结果Stack[0~top-1]为凸包的编号//一个点或两个点 则凸包为一或二个点int Stack[MAXN],top;Point vertex[MAXN];bool Graham_cmp(Point A, Point B){ double tmp=(A-vertex[0])^(B-vertex[0]); if(sgn(tmp) > 0) return 1; if(sgn(tmp) == 0 && sgn(dist(A,vertex[0])-dist(B,vertex[0])) <= 0) return 1; return 0;}
void Graham(int n){ int k=0; for(int i=1; i<n; i++) if((vertex[k].y>vertex[i].y) || (vertex[k].y==vertex[i].y && vertex[k].x>vertex[i].x)) k=i; swap(vertex[0], vertex[k]); sort(vertex+1, vertex+n, Graham_cmp); if(n == 1) { top=1; Stack[0]=0; return; } if(n == 2) { top=2; Stack[0]=0; Stack[1]=1; return; } Stack[0]=0; Stack[1]=1; top=2; for(int i=2; i<n; i++) { while(top > 1 && sgn((vertex[Stack[top-1]]-vertex[Stack[top-2]])^(vertex[i]-vertex[Stack[top-2]])) <= 0) top--; Stack[top++]=i; }}int main(){ int n,l; while(scanf("%d%d",&n,&l)!=EOF) { for(int i=0; i<n; i++) vertex[i].input(); Graham(n); double ans=0.0; for(int i=0; i<top; i++) ans+=dist(vertex[Stack[i]],vertex[Stack[(i+1)%top]]); ans+=2*PI*l; printf("%.f\n",ans); } return 0;}
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