2014
01-04

Maple trees

There are a lot of trees in HDU. Kiki want to surround all the trees with the minimal required length of the rope . As follow,

To make this problem more simple, consider all the trees are circles in a plate. The diameter of all the trees are the same (the diameter of a tree is 1 unit). Kiki can calculate the minimal length of the rope , because it’s so easy for this smart girl.
But we don’t have a rope to surround the trees. Instead, we only have some circle rings of different radius. Now I want to know the minimal required radius of the circle ring. And I don’t want to ask her this problem, because she is busy preparing for the examination.
As a smart ACMer, can you help me ?

The input contains one or more data sets. At first line of each input data set is number of trees in this data set n （1 <= n <= 100）, it is followed by n coordinates of the trees. Each coordinate is a pair of integers, and each integer is in [-1000, 1000], it means the position of a tree’s center. Each pair is separated by blank.
Zero at line for number of trees terminates the input for your program.

The input contains one or more data sets. At first line of each input data set is number of trees in this data set n （1 <= n <= 100）, it is followed by n coordinates of the trees. Each coordinate is a pair of integers, and each integer is in [-1000, 1000], it means the position of a tree’s center. Each pair is separated by blank.
Zero at line for number of trees terminates the input for your program.

2
1 0
-1 0
0

1.50

(思路明显错了)

// 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;
}

1. 在方法1里面：

//遍历所有的边，计算入度
for(int i=0; i<V; i++)
{
degree = 0;