http://poj.org/problem?id=2187

題意：

最長的點對近距離的平方：

題解：

?旋轉卡殼法， 要注意的地方是，有 所有點共線的情況，所以，（求凸包時）要將，共線點去出 ；

?

??1?#include<cstdio>
??2?#include<cstring>
??3?#include<cmath>
??4?#include<iostream>
??5?#include<algorithm>
??6?#include<set>
??7?#include<map>
??8?#include<queue>
??9?#include<vector>
?10?#include<string>
?11?#define?Min(a,b)?a<b?a:b
?12?#define?Max(a,b)?a>b?a:b
?13?#define?CL(a,num)?memset(a,num,sizeof(a));
?14?#define?maxn??50100
?15?#define?eps??1e-12
?16?#define?inf?100000000
?17?#define?mx?1<<60
?18?#define?ll???__int64
?19?const?double?pi??=?acos(-1.0);
?20?using?namespace?std;
?21?struct?point
?22?{
?23?????double?x,y;
?24?}p[maxn];
?25?point?stack[maxn];
?26?int?dblcmp(double?x)
?27?{
?28?????if(fabs(x)?<?eps)?return?0;
?29?????if(x?<?0)?return?-1;
?30?????else?return?1;
?31?}
?32?double?det(double?x1,double?y1,double?x2,double?y2)
?33?{
?34?????return?x1*y2?-?x2*y1?;
?35?}
?36?double?cross(point?a,?point?b,?point?c)
?37?{
?38?????return?det(b.x?-?a.x,?b.y?-?a.y,?c.x?-?a.x,?c.y?-?a.y);
?39?}
?40?int?cmp(point?a,point?b)
?41?{
?42?????if(a.y?!=?b.y)?return?a.y?<?b.y;
?43?????else?return?a.x?<?b.x;
?44?}
?45?int??top?,n;
?46?void?graham()
?47?{
?48?
?49?????int??i,len;
?50?????top?=?0;
?51?????sort(p,p+n,cmp);
?52?
?53?????if(n?==?0)?return?;
?54?????if(n?==?1)?return?;
?55?????stack[top++]?=?p[0];
?56?????stack[top++]?=?p[1];
?57?????for(i?=?2?;i<n;i++)//求右鏈
?58?????{
?59?????????while(top?>?1&&?cross(stack[top?-?1],stack[top?-?2],p[i])?>=?0)?top--;
?60?
?61?????????stack[top++]?=?p[i];
?62?????}
?63?
?64??????//dblcmp(cross(p[stack[top?-?1]],p[stack[top?-?2]],p[i]))?可以直接是?cross
?65?????len?=??top?;
?66?
?67?????for(i?=?n?-?2;i?>=?0;i--)//求左鏈
?68?????{
?69??????????while(top?>?len?&&?cross(stack[top?-?1],stack[top?-?2],p[i])?>=?0)top--;
?70??????????stack[top++]?=?p[i];
?71?
?72?????}
?73?????top--;//第一個點入棧兩次?所以?減?1
?74?
?75?}
?76?double?dis(point?a,point?b)
?77?{
?78???double?x1?=?a.x;
?79???double?y1?=?a.y;
?80???double?x2?=?b.x;
?81???double?y2?=?b.y;
?82???return?(x1?-?x2)*(x1?-?x2)+?(y1?-?y2)*(y1?-?y2);
?83?}
?84?double??rotating_calipers(point?stack[],int?top)
?85?{
?86?????int?i;
?87?????int?q??=?1;
?88?????double?ans?=?0;
?89????stack[top]?=?stack[0]?;
?90?
?91?
?92????for(i?=?0;i?<?top;i++?)
?93????{
?94????????while(cross(stack[i+1],stack[q+1],stack[i])?>?cross(stack[i?+?1],stack[q],stack[i]))
?95?????????q?=?(q+1)%top;
?96?
?97?
?98?
?99????????ans?=?max(ans,max(dis(stack[i],stack[q]),dis(stack[i+1],stack[q+1])));
100?
101????}
102????return?ans?;
103?}
104?int?main()
105?{
106?
107?
108???????int??i,s,j;
109?????//freopen("data.txt","r",stdin);
110?????while(scanf("%d",&n)!=EOF)
111?????{
112?????????for(i?=?0;i<?n;i++)
113?????????{
114?????????????scanf("%lf%lf",&p[i].x,&p[i].y);
115?????????}
116?????????graham()?;
117?
118????????double?ans?=?rotating_calipers(stack,top);
119????????printf("%.0lf\n",ans);
120?
121?
122?????}
123?}

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轉載于:https://www.cnblogs.com/acSzz/archive/2012/08/28/2660509.html

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