[BZOJ1670][Usaco2006 Oct]Building the Moat护城河的挖掘（凸包

题目描述

传送门

题解

凸包裸题。

代码

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define N 5005

const double eps=1e-9;
int dcmp(double x)
{
    if (x<=eps&&x>=-eps) return 0;
    return (x>0)?1:-1;
}
struct Vector
{
    double x,y;
    Vector(double X=0,double Y=0)
    {
        x=X,y=Y;
    }
    bool operator < (const Vector &a) const
    {
        return x<a.x||(x==a.x&&y<a.y);
    }
};
typedef Vector Point;
Vector operator - (Vector a,Vector b) {return Vector(a.x-b.x,a.y-b.y);}
int n,top;
Point p[N],stack[N];
double ans;

double Dot(Vector a,Vector b)
{
    return a.x*b.x+a.y*b.y;
}
double Len(Vector a)
{
    return sqrt(Dot(a,a));
}
double Cross(Vector a,Vector b)
{
    return a.x*b.y-a.y*b.x;
}
void graham()
{
    sort(p+1,p+n+1);
    top=0;
    for (int i=1;i<=n;++i)
    {
        while (top>1&&dcmp(Cross(stack[top]-stack[top-1],p[i]-stack[top-1]))<=0)
            --top;
        stack[++top]=p[i];
    }
    int k=top;
    for (int i=n-1;i>=1;--i)
    {
        while (top>k&&dcmp(Cross(stack[top]-stack[top-1],p[i]-stack[top-1]))<=0)
            --top;
        stack[++top]=p[i];
    }
    if (n>1) --top;
}

int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;++i) scanf("%lf%lf",&p[i].x,&p[i].y);
    graham();
    for (int i=1;i<=top;++i)
        ans+=Len(stack[i%top+1]-stack[(i+1)%top+1]);
    printf("%.2lfn",ans);
}

（编辑：厦门网）

【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容!