ACM学习历程—FZU 2140 Forever 0.5（计算几何 && 构造）

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

3

2

3

5

Sample Output

No

No

Yes

0.000000 0.525731

-0.500000 0.162460

-0.309017 -0.425325

0.309017 -0.425325

0.500000 0.162460

Hint

This problem is special judge.

 

题目大意就是找n个点满足上面的条件。

然而1、2、3个点显然不满足。

然后4个点的时候排除正方形，只能画出下面这种图形满足条件：

然后，发现，若需加入第五个点，制约条件3要求下，第5个点仅能与四个点中一个点距离为1。

然后由于其余点距离范围的控制加上面积的控制。我可以将第五个点选在距离A很近的，而且在以B为圆心1为半径的弧AD上。

由于需要在AD弧内能放下100个点，所以每次远离A点水平距离0.001。这样100个点只有0.1。然而AD水平距离为0.5，所以0.001到0.005内都是可以的。

而如果取到0.0001这样小，由于精度只有10^-6次方，所以在计算距离的时候平方会导致精度丢失而使新点几乎和A点效果一致，导致条件三不满足。

然后由于原四边形面积是0.5，如果算上整个扇形的面积是0.5 + PI/6 - sqrt(3)/4 < 0.75。所以面积满足。

由于整个图形是在一个直径为1的圆内，距离也满足。

 

代码:

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <string>
#define eps 0.001

using namespace std;

double x[] = {       0, 0.5,         0, -0.5};
double y[] = {0.866025,   0, -0.133975,    0};

inline double pow2(double a)
{
    return a*a;
}

int n;

void Work()
{
    double xx, yy;
    xx = x[0];
    for (int i = 0; i < 4; ++i)
        printf("%.6lf %.6lf\n", x[i], y[i]);
    n -= 4;
    xx -= eps;
    for (int i = 0; i < n; ++i)
    {
        yy = sqrt(1-pow2(xx-x[1]));
        printf("%.6lf %.6lf\n", xx, yy);
        xx -= eps;
    }
}

int main()
{
    //freopen("test.in", "r", stdin);
    int T;
    scanf("%d", &T);
    for (int times = 0; times < T; ++times)
    {
        scanf("%d", &n);
        if (n < 4)
            printf("No\n");
        else
        {
            printf("Yes\n");
            Work();
        }
    }
    return 0;
}