Codeforces Round #432 (Div. 2) C 850A Five Dimensional Points(思维)

我要承认这题我真的一点思路都没有.
想着怎么样能弄一个n^2logn的算法出来,还傻乎乎算了好久角啊边啊什么的,好菜好蠢啊.可能自从高中毕业之后就很久没接触几何了,有点懵逼.
这题的核心在于,假设二维平面上,一个点要和其余的点都构不成锐角(以该店为顶点),那么我们可以想象到,其余的点最多有四个,想一想,对吧?也就是一个十字架的情况.那么我们推广一下,三维中,最多的其余点只有6个,对吧?那么我们就猜想,五维中,最多10个.
也就是说,在五维空间中,只要点的总数>=11,那么就不可能有某一个点,是”good”的,这就是本题关键了.
所以我们可以对n>=11直接输出,而其余情况则进行暴力.

/*  xzppp  */#include <iostream>#include <vector>#include <cstdio>#include <string.h>#include <algorithm>#include <queue>#include <map>#include <string>#include <cmath>#include <bitset>#include <iomanip>using namespace std;#define FFF freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define MP make_pair#define PB push_backtypedef long long  LL;typedef unsigned long long ULL;typedef pair<int,int > pii;typedef pair<double,double > pdd;typedef pair<double,int > pdi;const int MAXN = 1000+17;const int MAXM = 20;const int MAXV = 2*1e3+17;const int INF = 0x7fffffff;const int MOD = 1e9+7;struct point{    int a,b,c,d,e;}all[MAXN];int n;bool check(int p){    bool can = true;    for (int i = 0; i < n; ++i)    {        if(i==p) continue;        for (int j = 0; j < n; ++j)        {            if(j==p||j==i) continue;            point t1,t2;            t1.a = all[i].a - all[p].a;            t1.b = all[i].b - all[p].b;            t1.c = all[i].c - all[p].c;            t1.d = all[i].d - all[p].d;            t1.e = all[i].e - all[p].e;            t2.a = all[j].a - all[p].a;            t2.b = all[j].b - all[p].b;            t2.c = all[j].c - all[p].c;            t2.d = all[j].d - all[p].d;            t2.e = all[j].e - all[p].e;            LL fz = t1.a*t2.a+t1.b*t2.b+t1.c*t2.c+t1.d*t2.d+t1.e*t2.e;            if(fz>0) can = false;        }    }    return can;}int main(){    #ifndef ONLINE_JUDGE     FFF    #endif    cin>>n;    for (int i = 0; i < n; ++i)    {        cin>>all[i].a>>all[i].b>>all[i].c>>all[i].d>>all[i].e;    }    if(n>11)        cout<<"0"<<endl;    else    {        vector<int > ans;        int cnt = 0;        for (int i = 0; i < n; ++i)            if(check(i)) ans.PB(i);        cout<<ans.size()<<endl;        for (int i = 0; i < ans.size(); ++i)        {            cout<<ans[i]+1<<endl;        }    }    return 0;}