| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 35396 | Accepted: 10961 |
Description
Bessie, Farmer John’s prize cow, has just won first place in a bovine beauty contest, earning the title ‘Miss Cow World’. As a result, Bessie will make a tour of N (2 <= N <= 50,000) farms around the world in order to spread goodwill between farmers and their cows. For simplicity, the world will be represented as a two-dimensional plane, where each farm is located at a pair of integer coordinates (x,y), each having a value in the range -10,000 … 10,000. No two farms share the same pair of coordinates.Even though Bessie travels directly in a straight line between pairs of farms, the distance between some farms can be quite large, so she wants to bring a suitcase full of hay with her so she has enough food to eat on each leg of her journey. Since Bessie refills her suitcase at every farm she visits, she wants to determine the maximum possible distance she might need to travel so she knows the size of suitcase she must bring.Help Bessie by computing the maximum distance among all pairs of farms.
Input
* Line 1: A single integer, N* Lines 2..N+1: Two space-separated integers x and y specifying coordinate of each farm
Output
Sample Input
4
0 0
0 1
1 1
1 0
Sample Output
2
题意：给出一些点，求平面最远点对距离的平方。
思想：我们知道，平面中最远点对一定是存在于这些点所围成的凸包中，那么，我们首先用Graham’s Scan算法求出凸包中的所有点，然后可以用O(n)的旋转卡壳来解决剩下的问题(当然暴力枚举也可以啦)。具体凸包的算法可以找我其他的文章哦！
AC代码：
#include<iostream>
#include<algorithm>
#include<cmath>
#include<stdio.h>
using namespace std;
#define max(a,b) (a>b?a:b)
#define PI acos(-1)
struct point
{
double x;
double y;
double distance(const point &b)const //计算距离
{
return (x-b.x)*(x-b.x)+(y-b.y)*(y-b.y);
}
point operator-(point s)
{
return {x-s.x,y-s.y};
}
} ;
point a[55000],data[55000];
int crossLeft(point p0,point p1,point p2) //判读左转还是右转
{
return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);
}
bool cmp(point p1,point p2) //极角序
{
int tmp=crossLeft(a[0],p1,p2);
if(tmp>0) return true;
else if(tmp==0&&a[0].distance(p1)<a[0].distance(p2))return true;
else return false;
}
bool cmp_x(point p1,point p2) //坐标序
{
if(p1.x!=p2.x)return p1.x<p2.x;
return p1.y<p2.y;
}
void init(int n)
{
int i,k=0;
point p;
scanf("%lf%lf",&a[0].x,&a[0].y);
p=a[0];
for(i=1; i<n; i++)
{
scanf("%lf%lf",&a[i].x,&a[i].y);
if( (p.y>a[i].y) || ((p.y==a[i].y)&&(p.x>a[i].x)) )
{
p=a[i];
k=i;
}
}
a[k]=a[0];
a[0]=p;
sort(a+1,a+n,cmp);
}
int main()
{
int N;
while(~scanf("%d",&N))
{
init(N);
for(int i=0; i<2; i++)
data[i]=a[i]; //前两个点入栈
int top=1;
for(int i=2; i<N; i++)
{
data[++top]=a[i]; //新点进栈
while(top>1&&crossLeft(data[top-2],data[top-1],data[top])<=0) //向后遍历判断是否存在凹的区域
data[top-1]=data[top],top--; //存在，出栈一个点
}
++top;
/* 旋转卡壳部分 */
if(top==2)printf("%.0f\n",data[0].distance(data[1])); //凸包退化的情况
else
{
double maxx=0.0;
int i=0,j=0; //某方向上的对踵点对
for(int k=0; k<top; k++)
{
if(!cmp_x(data[i],data[k]))i=k;
if(cmp_x(data[j],data[k]))j=k;
}
int si=i,sj=j;
while(i!=sj||j!=si) //旋转180°
{
maxx=max(maxx,data[i].distance(data[j]));
if(crossLeft({0,0},data[(i+1)%top]-data[i],data[(j+1)%top]-data[j])<0)
i=(i+1)%top;
else j=(j+1)%top;
}
printf("%.0f\n",maxx);
}
/* END */
}
}
