USACO 5.1.1 Fencing the Cows 圈奶牛 题解与分析

Fencing the Cows圈奶牛

译 by Felicia Crazy

目录

 [隐藏] 

• 1描述
• 2格式
• 3SAMPLE INPUT (file fc.in)
• 4SAMPLE OUTPUT (file fc.out)

描述

农夫约翰想要建造一个围栏用来围住他的奶牛，可是他资金匮乏。他建造的围栏必须包括他的奶牛喜欢吃草的所有地点。对于给出的这些地点的坐标，计算最短的能够围住这些点的围栏的长度。

格式

PROGRAM NAME: fc

INPUT FORMAT

输入数据的第一行包括一个整数 N。N（0 <= N <= 10,000）表示农夫约翰想要围住的放牧点的数目。接下来 N 行，每行由两个实数组成，Xi 和 Yi,对应平面上的放牧点坐标（-1,000,000 <= Xi,Yi <= 1,000,000）。数字用小数表示。

OUTPUT FORMAT

输出必须包括一个实数，表示必须的围栏的长度。答案保留两位小数。

SAMPLE INPUT (file fc.in)

44 84 125 9.37 8

SAMPLE OUTPUT (file fc.out)

12.00

【分析】：

      求二维凸包，计算几何问题可以参见：计算几何之基础知识及例题  ， 计算几何之题目讲解 

【代码】：

/*   ID:csyzcyj1   PROG:fc   LANG:C++*/#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>#include<iostream>#include<cmath>using namespace std;#define MAXN 20001struct point{double x,y;};point a[MAXN];int N,stack[MAXN],top;double ans=0.0;double dist(double x1,double y1,double x2,double y2){return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}bool cmp(point X,point Y){      double x1=X.x,y1=X.y,x2=Y.x,y2=Y.y;      if(x1*y2-x2*y1==0.0)            return x1<x2;      return ((x1*y2-x2*y1)>0.0);}void prepare(){      int basic=1;      for(int i=2;i<=N;i++)            if(a[i].x<a[basic].x || (a[i].x==a[basic].x && a[i].y<a[basic].y))                  basic=i;      swap(a[1],a[basic]);      for(int i=2;i<=N;i++)            a[i].x-=a[1].x,a[i].y-=a[1].y;      a[1].x=0.0,a[1].y=0.0;      sort(a+2,a+1+N,cmp);      a[N+1].x=0.0,a[N+1].y=0.0;}void Convexhull(){      int now1,now2;      double x1,x2,y1,y2;      top=3;      stack[1]=1,stack[2]=2,stack[3]=3;      for(int i=4;i<=N+1;i++)      {            now1=stack[top-1],now2=stack[top];            x1=a[now2].x-a[now1].x;            y1=a[now2].y-a[now1].y;            x2=a[i].x-a[now2].x;            y2=a[i].y-a[now2].y;                 while((x1*y2-x2*y1)<0.0)            {                  top--;                  now1=stack[top-1],now2=stack[top];                  x1=a[now2].x-a[now1].x;                  y1=a[now2].y-a[now1].y;                  x2=a[i].x-a[now2].x;                  y2=a[i].y-a[now2].y;            }             stack[++top]=i;            }}void figure(){      for(int i=top;i>=2;i--)            ans+=dist(a[stack[i]].x,a[stack[i]].y,a[stack[i-1]].x,a[stack[i-1]].y);}int main(){      freopen("fc.in","r",stdin);  freopen("fc.out","w",stdout);   scanf("%d",&N);  for(int i=1;i<=N;i++)        scanf("%lf%lf",&a[i].x,&a[i].y);      prepare();      Convexhull();      figure();      printf("%.2lf\n",ans);  //system("pause");      return 0;}

 

测评信息：

TASK: fcLANG: C++Compiling...Compile: OKExecuting...   Test 1: TEST OK [0.000 secs, 3772 KB]   Test 2: TEST OK [0.000 secs, 3772 KB]   Test 3: TEST OK [0.011 secs, 3772 KB]   Test 4: TEST OK [0.011 secs, 3772 KB]   Test 5: TEST OK [0.011 secs, 3772 KB]   Test 6: TEST OK [0.011 secs, 3772 KB]   Test 7: TEST OK [0.022 secs, 3772 KB]   Test 8: TEST OK [0.065 secs, 3772 KB]All tests OK.YOUR PROGRAM ('fc') WORKED FIRST TIME!  That's fantastic-- and a rare thing.  Please accept these special automatedcongratulations.

转载注明出处：http://blog.csdn.net/u011400953