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POJ 1113 凸包模板题

时间:2014-08-15 19:26:09      阅读:271      评论:0      收藏:0      [点我收藏+]

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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;
}
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; }

  

POJ 1113 凸包模板题,布布扣,bubuko.com

POJ 1113 凸包模板题

标签:blog   os   io   for   ar   div   amp   log   

原文地址:http://www.cnblogs.com/Mathics/p/3915435.html

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