最大三角形 HDU

来源：互联网发布：商业数据分析模型编辑：程序博客网时间：2024/05/19 23:59
整理了一下旋转卡壳的模板，把接口统一了一下
//#include<bits/stdc++.h>  //#pragma comment(linker, "/STACK:1024000000,1024000000")   #include<stdio.h>  #include<algorithm>  #include<queue>  #include<string.h>  #include<iostream>  #include<math.h>  #include<set>  #include<map>  #include<vector>  #include<iomanip>  using namespace std;    const double pi=acos(-1.0);  #define ll long long  #define pb push_back#define sqr(a) ((a)*(a))#define dis(a,b) sqrt(sqr(a.x-b.x)+sqr(a.y-b.y))const double eps=1e-6;const int maxn=5e4+56;const int inf=0x3f3f3f3f;int n;int tot;//凸包上点数struct Point{double x,y;Point(){}Point(double x,double y):x(x),y(y){}}point[maxn],vertex[maxn];bool cmp(Point a,Point b){return(a.y<b.y||(a.y== b.y && a.x<b.x));}double xmult(Point p1,Point p2,Point p3){//p3p1,p3p2的夹角测试return ( (p1.x-p3.x)*(p2.y-p3.y)-(p1.y-p3.y)*(p2.x-p3.x) );}//正表示p1在p2的顺时针方向int Andrew(){//返回凸包顶点数sort(point,point+n,cmp);int top=1;vertex[0]=point[0];vertex[1]=point[1];for(int i=2;i<n;i++){ while(top && xmult(point[i],vertex[top],vertex[top-1])>eps)top--;vertex[++top]=point[i];}int len=top;vertex[++top]=point[n-2];for(int i=n-3;i>=0;i--){while(top!=len && xmult(point[i],vertex[top],vertex[top-1])>eps)top--;vertex[++top]=point[i];}return top;}double rotating_calipers_r(int n){//返回最大直径int q=1;double ans=0;vertex[n]=vertex[0];for(int p=0;p<n;p++){while(xmult(vertex[p+1],vertex[q+1],vertex[p])>xmult(vertex[p+1],vertex[q],vertex[p]))q=(q+1)%n;ans=max(ans,max(dis(vertex[p],vertex[q]),dis(vertex[p+1],vertex[q+1])));//更新方法是为了处理p,p+1 // q,q+1这种平行状况}return ans;}double rotating_calipers_s(int m){//最大三角形面积if(m==1||m==2)return 0;double ans=0;for(int i=0;i<m;i++){int q=1;for(int j=i+1;j<m;j++){while(xmult(vertex[j],vertex[q+1],vertex[i])>xmult(vertex[j],vertex[q],vertex[i]))q=(q+1)%m;ans=max(ans,xmult(vertex[i],vertex[j],vertex[q]));}}return ans/2.0;}int main(){while(~scanf("%d",&n)){for(int i=0;i<n;i++){scanf("%lf%lf",&point[i].x,&point[i].y);}int m=Andrew();printf("%.2lf\n",rotating_calipers_s(m));}}
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