Accept: 269 Submit: 934 Special Judge
Time Limit: 1000 mSec Memory Limit : 32768 KB
Problem Description
Given an integer N, your task is to judge whether there exist N points in the plane such that satisfy the following conditions:
1. The distance between any two points is no greater than 1.0.
2. The distance between any point and the origin (0,0) is no greater than 1.0.
3. There are exactly N pairs of the points that their distance is exactly 1.0.
4. The area of the convex hull constituted by these N points is no less than 0.5.
5. The area of the convex hull constituted by these N points is no greater than 0.75.
Input
The first line of the date is an integer T, which is the number of the text cases.
Then T cases follow, each contains an integer N described above.
1 <= T <= 100, 1 <= N <= 100
Output
For each case, output “Yes” if this kind of set of points exists, then output N lines described these N points with its coordinate. Make true that each coordinate of your output should be a real number with AT MOST 6 digits after decimal point.
Your answer will be accepted if your absolute error for each number is no more than 10-4.
Otherwise just output “No”.
See the sample input and output for more details.
Sample Input
Sample Output
開始半天沒有讀懂題意,以爲這道題挺複雜的,後來讀懂了一點,原來也是比較水,有一點技巧。
題意:給你n個點,這n個點是否滿足下面的條件。
1.兩點之間的距離不大於1
2.任意點到原點的距離不大於1
3.有n對點的距離剛好等於1
4.n個點構成的n邊形的面積不小於0.5
5.n個點構成的n邊形的面積不大於0.75
如果滿足關係就輸出yes + 這n個點,沒有就輸出 no。
思路:畫個圖可以進行推敲一下,前四個點基本上都是確定的。3個點就是一個等邊三角形,(原點,x=1,y=0)但是面積不符合,所以滿足條件至少需要4個點,其實就是一個半徑爲1的圓以原點爲圓心。然後以原點和x軸畫一個邊長爲1的等邊三角形,這樣的話就有三個點了,然後剩下的點從圓上找就可以了,主要是離那三個點的距離大於1即可,因爲圓上的點到圓心的距離都爲1,其實就是將圓離散化。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
using namespace std;
double x[110],y[110];
const double temp=0.001;
void solve()
{
x[0]=0,y[0]=0;//原點
x[1]=1,y[1]=0;//第二個點
x[2]=0.5,y[2]=sqrt(1-0.25);//等邊三角形的三個點
x[3]=0.5,y[3]=y[2]-1;
for(int i=4;i<110;i++)
{
y[i]=i*temp;//離散化
x[i]=sqrt(1-y[i]*y[i]);
}
}
int main()
{
int t,n;
solve();
cin>>t;
while(t--)
{
cin>>n;
if(n<4)
printf("No\n");
else
{
printf("Yes\n");
for(int i=0;i<n;i++)
printf("%.6lf %.6lf\n",y[i],x[i]);
}
}
return 0;
}