hdu 2215 (最小覆盖圆)(目前WA)

(思路明显错了)

最小覆盖圆必定是某三个顶点的外接圆。

先求凸包再枚举。

 

// hdu 2215.cpp : Defines the entry point for the console application. // #include "stdafx.h" #include "math.h" #include <iostream> using namespace std; #define ABS_N(x) ((x)<0?-(x):(x)) int N,stack_top; struct Node { int x,y; }m_stack[1005],position[1005]; inline int CrossMutiply(Node p1,Node p2,Node p3) { return (p2.x-p1.x)*(p3.y-p1.y)-(p2.y-p1.y)*(p3.x-p1.x); } inline int Distance(Node p1,Node p2) { return (p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y); } int CMP(const void* a,const void* b) { Node *p1=(Node*)a,*p2=(Node*)b; int m=CrossMutiply(position[0],*p1,*p2); if(m==0) return Distance(position[0],*p1)-Distance(position[0],*p2); else return -m; } void Convex() { int swap=0; for(int i=1;i<N;i++) { if((position[i].y<position[swap].y)||(position[i].y==position[swap].y&&position[i].x<position[swap].x)) swap=i; } Node temp=position[swap]; position[swap]=position[0]; position[0]=temp; qsort(position+1,N-1,sizeof(position[0]),CMP); m_stack[0]=position[0]; m_stack[1]=position[1]; stack_top=1; for(int i=2;i<N;i++) { while(stack_top>=1&&CrossMutiply(m_stack[stack_top-1],m_stack[stack_top],position[i])<=0) stack_top--; m_stack[++stack_top]=position[i]; } } inline double Diameter(Node p1,Node p2,Node p3) { double temp=sqrt((double)Distance(p2,p3)); double temp2=CrossMutiply(p1,p2,p3); temp2=temp2/(sqrt((double)(Distance(p1,p2)*Distance(p1,p3)))); temp=temp/temp2; return ABS_N(temp); } int main() { freopen("d://1.txt","r",stdin); double ans,temp; while(scanf("%d",&N)&&N) { for(int i=0;i<N;i++) { scanf("%d%d",&position[i].x,&position[i].y); } if(N==1) { printf("%.2f/n",0.5); continue; } else if(N==2) { printf("%.2f/n",sqrt((double)Distance(position[0],position[1]))/2+0.5); continue; } Convex(); if(stack_top==1) { printf("%.2f/n",sqrt((double)Distance(m_stack[0],m_stack[1]))/2+0.5); continue; } ans=0; for(int i=0;i<stack_top-1;i++) { for(int j=i+1;j<stack_top;j++) { for(int k=j+1;k<=stack_top;k++) { temp=Diameter(m_stack[i],m_stack[j],m_stack[k]); if(temp>ans) ans=temp; } } } printf("%.2f/n",ans/2+0.5); } return 0; }