Codeforces Gym 100523K K - Cross Spider 计算几何，判断是否n点共面

K - Cross Spider
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87794#problem/K

Description

The Bytean cross spider (Araneida baitoida) is known to have an amazing ability. Namely, it can instantly build an arbitrarily large spiderweb as long as it is contained in a single plane. This ability gives the spider an opportunity to use a fancy hunting strategy. It does not need to wait until a fly is caught in an already built spiderweb; if only the spider knows the current position of a fly, it can instantly build a spiderweb to catch the fly. A cross spider has just spotted n flies in Byteasar’s garden. Each fly is flying still in some point of a 3D space. The spider is wondering if it can catch all the flies with a single spiderweb. Write a program that answers the spider’s question.

Input

The first line of the input contains an integer n (1 ¬ n ¬ 100 000). The following n lines contain a description of the flies in a 3D space: the i-th line contains three integers xi , yi , zi (−1 000 000 ¬ xi ; yi ; zi ¬ 1 000 000) giving the coordinates of the i-th fly (a point in a 3-dimensional Euclidean space). No two flies are located in the same point.

Output

Your program should output a single word TAK (i.e., yes in Polish) if the spider can catch all the flies with a single spiderweb. Otherwise your program should output the word NIE (no in Polish).

Sample Input

4 0 0 0 -1 0 -100 100 0 231 5 0 15

Sample Output

TAK

HINT

 

题意

一个三维空间，给n个点，问着n个点是否在同一个平面上

题解：

首先先找到不在同一条直线上的三个点，做法向量，然后我们再枚举任意两个点，如果这两个点是和法向量垂直的话，就说明在一个平面上

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <iomanip>
#include <string>
#include <ctime>
#include <list>
#include <bitset>
typedef unsigned char byte;
#define pb push_back
#define input_fast std::ios::sync_with_stdio(false);std::cin.tie(0)
#define local freopen("in.txt","r",stdin)
#define pi acos(-1)

using namespace std;

typedef struct point
{
  long long x , y , z;
};


const int maxn = 1e5 + 500;
const double eps = 1e-7;
int n ;
point A[maxn];
point vi;

bool judge(point a,point b)
{
    return a.x * b.y == b.x * a.y && a.y*b.z == b.y*a.z && a.x * b.z == a.z*b.x;
}


int main(int argc,char *argv[])
{
    scanf("%d",&n);
    for(int i = 0 ; i < n ; ++ i) scanf("%I64d%I64d%I64d",&A[i].x,&A[i].y,&A[i].z);
    if (n <= 3)
    {
         printf("TAK\n");
         return 0;
    }
    else
    {
        int ok = 1;
        for(int i = 0 ; i < n ; ++ i)
        {
            int a = i ;
            int b = (i+1) % n;
            int c = (i+2) % n;
            point vi1 , vi2;
            vi1.x = A[b].x - A[a].x;
            vi1.y = A[b].y - A[a].y;
            vi1.z = A[b].z - A[a].z;
            vi2.x = A[c].x - A[b].x;
            vi2.y = A[c].y - A[b].y;
            vi2.z = A[c].z - A[b].z;
            if (judge(vi1,vi2)) continue;
            else
            {
                vi.x = vi1.y*vi2.z - vi2.y*vi1.z;
                vi.y = vi2.x*vi1.z - vi1.x*vi2.z;
                vi.z = vi1.x*vi2.y - vi1.y*vi2.x;
                ok = 0;
                break;
            }
        }
       if (ok)
       {
           printf("TAK\n");
           return 0;
       }
       else
       {
             ok = 1;
             for(int i = 0 ; i < n ; ++ i)
             {
                 int a = i ;
                 int b = (i+1) % n;
                 point tx;
                 tx.x =  A[b].x - A[a].x;
                 tx.y =  A[b].y - A[a].y;
                 tx.z =  A[b].z - A[a].z;
                 if (vi.x * tx.x + vi.y * tx.y + vi.z * tx.z != 0)
                 {
                     ok = 0;
                     break;
                 }
             }
       }
       if (ok)     printf("TAK\n");
       else printf("NIE\n");
       return 0;
    }
    return 0;
}