POJ 2187.Beauty Contest

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题目:http://poj.org/problem?id=2187

AC代码(C++):

#include <iostream>#include <algorithm>#include <stdio.h>#include <vector>#include <queue>#include <math.h>#include <string>#include <string.h>#include <bitset>#define INF 0x7fffffff#define LFINF 1e10#define eps 1e-7#define MAXN 100105#define PI 3.14159265358979323846using namespace std;struct Point  {      double x,y;    Point(double x=0,double y=0):x(x),y(y) {}};typedef Point Vector;Vector operator - (Vector A,Vector B) { return Vector(A.x-B.x,A.y-B.y); }int dcmp(double x){      if(fabs(x)<eps) return 0;      else return x<0?-1:1;  }int n;Point node[50005];Point convex[50005];int top;double Cross(Vector A,Vector B) { return (A.x*B.y-A.y*B.x); }double Dot(Vector A,Vector B) { return A.x*B.x+A.y*B.y; }  double Length(Vector A) { return sqrt(Dot(A,A)); }  bool cmp(Point a, Point b){    return dcmp(a.y-b.y)==-1 || (dcmp(a.y-b.y)==0 && dcmp(a.x-b.x)==-1);}void grahamScan(){sort(node,node+n,cmp);    int tmp;    top = 1;    convex[0]=node[0], convex[1]=node[1];    for(int i=2; i<n; i++){        while(top>0 && dcmp(Cross(convex[top]-convex[top-1], node[i]-convex[top-1])) <= 0)top--;        convex[++top] = node[i];    }    tmp = top;    for(int i=n-2; i>=0; i--){        while(top>tmp && dcmp(Cross(convex[top]-convex[top-1], node[i]-convex[top-1])) <= 0)top--;        convex[++top] = node[i];    }}double rotating_calipers(){      int q = 1;    double ans = 0;    convex[top] = convex[0];    for(int p = 0; p < top; p++)      {          while(dcmp(Cross(convex[p+1]-convex[p],convex[q+1]-convex[p])-Cross(convex[p+1]-convex[p],convex[q]-convex[p]))==1)            q=(q+1)%top;        ans=max(ans,max(Length(convex[p]-convex[q]),Length(convex[p+1]-convex[q+1])));    }      return ans;   }int main(){cin>>n;for(int i = 0; i < n; i++)cin>>node[i].x>>node[i].y;grahamScan();double ans = rotating_calipers();ans *= ans;printf("%.0lf",ans);}

总结: 凸包. 先计算凸包, 然后用旋转卡壳法计算凸包直径.

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