hdu 5533 正n边形判断 精度处理

Dancing Stars on Me

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1098    Accepted Submission(s): 598

Problem Description

The sky was brushed clean by the wind and the stars were cold in a black sky. What a wonderful night. You observed that, sometimes the stars can form a regular polygon in the sky if we connect them properly. You want to record these moments by your smart camera. Of course, you cannot stay awake all night for capturing. So you decide to write a program running on the smart camera to check whether the stars can form a regular polygon and capture these moments automatically.

Formally, a regular polygon is a convex polygon whose angles are all equal and all its sides have the same length. The area of a regular polygon must be nonzero. We say the stars can form a regular polygon if they are exactly the vertices of some regular polygon. To simplify the problem, we project the sky to a two-dimensional plane here, and you just need to check whether the stars can form a regular polygon in this plane.

 

 

Input

The first line contains a integer T indicating the total number of test cases. Each test case begins with an integer n, denoting the number of stars in the sky. Following n lines, each contains 2 integers xi,yi, describe the coordinates of n stars.

1≤T≤300
3≤n≤100
−10000≤xi,yi≤10000
All coordinates are distinct.

 

 

Output

For each test case, please output "`YES`" if the stars can form a regular polygon. Otherwise, output "`NO`" (both without quotes).

 

 

Sample Input

3 3 0 0 1 1 1 0 4 0 0 0 1 1 0 1 1 5 0 0 0 1 0 2 2 2 2 0

 

 

Sample Output

NO YES NO

 

 

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

 

 

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题意：给你n个点(-1e4<x,y<=1e4)，判断这n个点能否组成一个正n边形；

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <algorithm>
#include <set>
#define MM(a) memset(a,0,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const double eps = 1e-12;
const int inf = 0x3f3f3f3f;
const double pi=acos(-1);
using namespace std;

struct Point{
    int x,y;
    void read()
    {
        scanf("%d%d",&x,&y);
    }
}p[105],tubao[105];

int dcmp(double a)
{
    if(fabs(a)<eps) return 0;
    else if(a>0) return 1;
    else return -1;
}

Point operator-(Point a,Point b)
{
    return (Point){a.x-b.x,a.y-b.y};
}

double dis(Point a)
{
    return sqrt(a.x*a.x+a.y*a.y);
}

double cross(Point a,Point b)
{
    return  a.x*b.y-b.x*a.y;
}

double dot(Point a,Point b)
{
    return a.x*b.x+a.y*b.y;
}

bool cmp(Point a,Point b)
{
    if(a.x!=b.x) return a.x<b.x;
    else return a.y<b.y;
}

int  convex_hull(Point *p,int n,Point *tubao)
{
    sort(p+1,p+n+1,cmp);
    int m=0;
    for(int i=1;i<=n;i++)
    {
        while(m>=2&&cross(p[i]-tubao[m-1],tubao[m]-tubao[m-1])>0) m--;
        tubao[++m]=p[i];
    }
    int k=m;
    for(int i=n-1;i>=1;i--)
    {
        while(m-k>=1&&cross(p[i]-tubao[m-1],tubao[m]-tubao[m-1])>0) m--;
        tubao[++m]=p[i];
    }
    m--;
    return m;
}

int main()
{
    int cas,n;
    scanf("%d",&cas);
    while(cas--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++) p[i].read();
        int k=convex_hull(p,n,tubao);
        tubao[k+1]=tubao[1];

        bool flag=true;
        double tmp=(n-2.0)*pi/n;

        for(int i=1;i<=k-1;i++)
        {
            Point a=tubao[i+1]-tubao[i],b=tubao[i+2]-tubao[i+1];
            double cosang=dot(a,b)/(dis(a)*dis(b));
            double ang=acos(cosang);
            ang=pi-ang;
            if(dcmp(ang-tmp)!=0) {flag=false;break;}
        }
        if(flag) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

　　分析：主要是借助这道题来分下下计算几何的精度问题，

double型数据精度处理的两种方式

1.相除改为ong long相乘，这种是肯定对的，不会错。

2.dcmp函数，这种比较简单，但是有一定的精度条件，如果角度是1/999999-1/1000000,那么相减起来就是1e-6*1/999999为1e-12级别，这样是可以使用dcmp的，比如本道题，因为1-e4<=x<=1e4，那么最小的角度差是1/（2*1e4-1）-1/2*1e4（最小的角是1/2*1e4，第二小的角度是1/（2*1e4-1））为1e-8级别>1e-12级别，所以可以用dcmp（eps<1e-12）