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 1005 using 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; }<br> 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; }
那么多的束缚,我不曾放弃过;那么多的险阻,我不曾倒下过。