题意:给定N个点,用矩形将所有点覆盖,要求矩形宽度最小。
思路:裸体,旋转卡壳去rotate即可。
最远距离是点到点;宽度是点到边。
#include<bits/stdc++.h> #define ll long long #define rep(i,a,b) for(int i=a;i<=b;i++) #define RC rotating_calipers using namespace std; const int maxn=400010; struct point{ ll x,y; point(double x=0,double y=0):x(x),y(y){} bool operator < (const point &c) const { return x<c.x||(x==c.x&&y<c.y);} point operator - (const point &c) const { return point(x-c.x,y-c.y);} }; ll det(point A,point B){ return A.x*B.y-A.y*B.x;} ll det(point O,point A,point B){ return det(A-O,B-O);} point a[maxn],ch[maxn]; void convexhull(int n,int &top) { sort(a+1,a+n+1); top=0; for(int i=1;i<=n;i++){ while(top>1&&det(ch[top-1],ch[top],a[i])<=0) top--; ch[++top]=a[i]; } int ttop=top; for(int i=n-1;i>=1;i--){ while(top>ttop&&det(ch[top-1],ch[top],a[i])<=0) top--; ch[++top]=a[i]; } } double rotating_calipers(point p[],int top) { double ans=123456789009876; int now=2; rep(i,1,top-1){ while(abs(det(p[i],p[i+1],p[now]))<abs(det(p[i],p[i+1],p[now+1]))){ now++; if(now==top) now=1; } double tmp=fabs(1.0*det(p[i],p[i+1],p[now])); tmp/=sqrt(1.0*(p[i].x-p[i+1].x)*(p[i].x-p[i+1].x)+(p[i].y-p[i+1].y)*(p[i].y-p[i+1].y)); ans=min(ans,tmp); } return ans; } int main() { int N; double S; scanf("%d%lf",&N,&S); for(int i=1;i<=N;i++) scanf("%I64d%I64d",&a[i].x,&a[i].y); int top; convexhull(N,top); printf("%.15lf\n",RC(ch,top)); return 0; }
原文地址:https://www.cnblogs.com/hua-dong/p/9623472.html
时间: 2024-10-15 04:23:03