zoukankan      html  css  js  c++  java
  • USACO Section 5.1 Fencing the Cows(凸包)

    裸的凸包..很好写,废话不说,直接贴代码。

    -----------------------------------------------------------------------------

    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    #include<cmath>
    #define rep(i,r) for(int i=0;i<r;i++)
    using namespace std;
    const int maxn=10000+5;
    struct P {
        
        double x,y;
        
        P(double x=0,double y=0):x(x),y(y) {}
        
        P operator + (P o) { return P(x+o.x,y+o.y); }
        P operator - (P o) { return P(x-o.x,y-o.y); }
        
        
        bool operator == (const P &o) const {
            return x==o.x && y==o.y;
        }
        bool operator < (const P &o) const {
            return x<o.x || (x==o.x && y<o.y);
        }
        
    };
    P ans[maxn],p[maxn];
    inline double cross(P a,P b) { return a.x*b.y-a.y*b.x; }
    inline double dist(P o) { return sqrt(o.x*o.x+o.y*o.y); }
    double convexHull(int n) {
        
        sort(p,p+n);
        
        int m=0;
        rep(i,n) {
            while(m>1 && cross(ans[m-1]-ans[m-2],p[i]-ans[m-2])<=0) m--;
            ans[m++]=p[i];
        }
        
        int k=m;
        for(int i=n-2;i>=0;i--) {
            while(m>k && cross(ans[m-1]-ans[m-2],p[i]-ans[m-2])<=0) m--;
            ans[m++]=p[i];
        }
        
        if(n>1) m--;
        
        double length=0;
        rep(i,m) length+=dist(ans[i]-ans[i+1]);
        
        return length;
        
    }
        
    int main()
    {
        freopen("fc.in","r",stdin); freopen("fc.out","w",stdout);
        
        
        int n;
        cin>>n;
        rep(i,n) scanf("%lf%lf",&p[i].x,&p[i].y);
        printf("%.2lf ",convexHull(n));
        
        
        return 0;
    }

    ----------------------------------------------------------------------------- 

    Fencing the Cows
    Hal Burch

    Farmer John wishes to build a fence to contain his cows, but he's a bit short on cash right. Any fence he builds must contain all of the favorite grazing spots for his cows. Given the location of these spots, determine the length of the shortest fence which encloses them.

    PROGRAM NAME: fc

    INPUT FORMAT

    The first line of the input file contains one integer, N. N (0 <= N <= 10,000) is the number of grazing spots that Farmer john wishes to enclose. The next N line consists of two real numbers, Xi and Yi, corresponding to the location of the grazing spots in the plane (-1,000,000 <= Xi,Yi <= 1,000,000). The numbers will be in decimal format.

    SAMPLE INPUT (file fc.in)

    4
    4 8
    4 12
    5 9.3
    7 8
    

    OUTPUT FORMAT

    The output should consists of one real number, the length of fence required. The output should be accurate to two decimal places.

    SAMPLE OUTPUT (file fc.out)

    12.00
  • 相关阅读:
    MySQL线程独享内存参数
    asp.net使用飞信fetionAPI接口免费发送短信的c#的实例
    VS2008不能创建解决方案
    转载:nginx配置文件的location标签执行顺序和反向代理配置
    图文讲解如何使用Gmail绑定域名开通企业邮箱(使用时代互联的域名管理后台)
    如何在博客园上放google adsense广告,并且不被博客园屏蔽掉google adsense的src
    kingcms 标签
    ADO.NET与SQL Server数据库的交互
    asp.net结合jQuery实现google suggest效果
    ADO.NET与Sql Server和Access的连接
  • 原文地址:https://www.cnblogs.com/JSZX11556/p/4337879.html
Copyright © 2011-2022 走看看