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可以做网站的服务器,网络推广方法的分类,开源门户网站源码,网站建设利润越来越低计算几何/旋转卡壳学习旋转卡壳请戳这里~感觉讲的最好的就是这个了…… 其实就是找面积最大的三角形？。。。并且满足单调…… 嗯反正就是这样…… 这是一道模板题好像必须写成循环访问？我在原数组后面复制了一遍点，结果挂了……改成curcur%…

计算几何/旋转卡壳

　　学习旋转卡壳请戳这里~感觉讲的最好的就是这个了……

　　其实就是找面积最大的三角形？。。。并且满足单调……

　　嗯反正就是这样……

　　这是一道模板题

　　好像必须写成循环访问？我在原数组后面复制了一遍点，结果挂了……改成cur=cur%n+1就过了QAQ

//其实是不是数组没开够所以复制的方法就爆了？

UPD：（2015年5月13日 20:40:45）

　　其实是我点保存在1~n里面，所以复制的时候不能写p[i+n-1]=p[i]; 而应该是p[i+n]=p[i];……QAQ我是傻逼

 1 Source Code
 2 Problem: 2187        User: sdfzyhy
 3 Memory: 1044K        Time: 32MS
 4 Language: G++        Result: Accepted
 5 
 6     Source Code
 7 
 8     //POJ 2187
 9     #include<cstdio>
10     #include<cstring>
11     #include<cstdlib>
12     #include<iostream>
13     #include<algorithm>
14     #define rep(i,n) for(int i=0;i<n;++i)
15     #define F(i,j,n) for(int i=j;i<=n;++i)
16     #define D(i,j,n) for(int i=j;i>=n;--i)
17     #define pb push_back
18     using namespace std;
19     typedef long long LL;
20     inline int getint(){
21         int r=1,v=0; char ch=getchar();
22         for(;!isdigit(ch);ch=getchar()) if (ch=='-') r=-1;
23         for(; isdigit(ch);ch=getchar()) v=v*10-'0'+ch;
24         return r*v;
25     }
26     const int N=100010;
27     /*******************template********************/
28     struct Poi{
29         int x,y;
30         Poi(){}
31         Poi(int x,int y):x(x),y(y){}
32         void read(){x=getint();y=getint();}
33     }p[N],ch[N];
34     typedef Poi Vec;
35     Vec operator - (const Poi &a,const Poi &b){return Vec(a.x-b.x,a.y-b.y);}
36     bool operator < (const Poi &a,const Poi &b){return a.x<b.x || (a.x==b.x && a.y<b.y);}
37     inline int Cross(const Poi &a,const Poi &b){return a.x*b.y-a.y*b.x;}
38     inline int Dot(const Poi &a,const Poi &b){return a.x*b.x+a.y*b.y;}
39     int n,m,ans;
40 
41     void graham(Poi *p,int n){
42         int size=0;
43         sort(p+1,p+n+1);
44         ch[++m]=p[1];
45         F(i,2,n){
46             while(m>1 && Cross(ch[m]-ch[m-1],p[i]-ch[m-1])<=0) m--;
47             ch[++m]=p[i];
48         }
49         int k=m;
50         D(i,n-1,1){
51             while(m>k && Cross(ch[m]-ch[m-1],p[i]-ch[m-1])<=0) m--;
52             ch[++m]=p[i];
53         }
54         if (n>1) m--;
55     }
56 
57     void rot(Poi *p,int n){
58         ans=0;
59     //    F(i,1,n-1) p[n+i-1]=p[i];
60     //    n=n*2-1;
61         int cur=2;
62         F(i,1,n){
63             Vec v = p[i]-p[i+1];
64             while(Cross(p[i+1]-p[i],p[cur+1]-p[i]) > Cross(p[i+1]-p[i],p[cur]-p[i]))
65                 cur=(cur%n)+1;
66             ans=max(ans,Dot(p[cur]-p[i],p[cur]-p[i]));
67             ans=max(ans,Dot(p[cur+1]-p[i+1],p[cur+1]-p[i+1]));
68         }
69     }
70                 
71     int main(){
72     #ifndef ONLINE_JUDGE
73         freopen("2187.in","r",stdin);
74     //    freopen("2187.out","w",stdout);
75     #endif
76         n=getint();
77         F(i,1,n) p[i].read();
78         graham(p,n);
79         rot(ch,m);
80         printf("%d\n",ans);
81         return 0;
82     }

View Code

Beauty Contest

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 29879		Accepted: 9260

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

* Line 1: A single integer that is the squared distance between the pair of farms that are farthest apart from each other.

Sample Input