标签:blog 使用 os io for ar div amp
旋 转卡壳水题。
直接使用旋转卡壳求距离。
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; const int MAXN=50100; struct point { int x,y; }p[MAXN]; int n; int st[MAXN],ans[MAXN]; int stop,cnt; bool cmp(point A,point B){ if(A.y<B.y) return true; else if(A.y==B.y){ if(A.x<B.x) return true; } return false; } int dist(point a,point b){ return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y); } bool multi(point a,point b,point c){ point p,q; p.x=a.x-c.x; p.y=a.y-c.y; q.x=b.x-c.x; q.y=b.y-c.y; return (p.x*q.y-q.x*p.y)>=0; } int TriangleArea(point a,point b,point c){ point p,q; p.x=a.x-c.x; p.y=a.y-c.y; q.x=b.x-c.x; q.y=b.y-c.y; return abs(p.x*q.y-q.x*p.y); } void slove(){ stop=cnt=0; st[stop++]=0; st[stop++]=1; for(int i=2;i<n;i++){ while(stop>1&&multi(p[i],p[st[stop-1]],p[st[stop-2]])) stop--; st[stop++]=i; } for(int i=0;i<stop;i++) ans[cnt++]=st[i]; stop=0; st[stop++]=n-1; st[stop++]=n-2; for(int i=n-3;i>=0;i--){ while(stop>1&&multi(p[i],p[st[stop-1]],p[st[stop-2]])) stop--; st[stop++]=i; } for(int i=1;i<stop;i++) ans[cnt++]=st[i]; } int for_dist(){ int distant=0; int q=1; for(int i=0;i<cnt-1;i++){ while(TriangleArea(p[ans[i]],p[ans[i+1]],p[ans[q+1]])> TriangleArea(p[ans[i]],p[ans[i+1]],p[ans[q]])) q=(q+1)%cnt; distant=max(distant,max(max(dist(p[ans[i]],p[ans[q]]),dist(p[ans[i+1]],p[ans[q+1]])),dist(p[ans[i+1]],p[ans[q]]))); } return distant; } int main(){ while(scanf("%d",&n)!=EOF){ for(int i=0;i<n;i++){ scanf("%d%d",&p[i].x,&p[i].y); } sort(p,p+n,cmp); slove(); printf("%d\n",for_dist()); } return 0; }
标签:blog 使用 os io for ar div amp
原文地址:http://www.cnblogs.com/jie-dcai/p/3888445.html