Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017) C. Five Dimensional Points

题意：给出一些五维坐标里的点，问有多少个好点，好点的定义为与其他任意2个点的向量夹角>=90°

思路：枚举，当这个点为好点的时候，我们可以确定另外2个点必为坏点，根据三角形内角和180°，然后跑就行

#include <bits/stdc++.h>
using namespace std;

const int N = 1e3+100;
int ans[N],vis[N],n;
struct node{
    int a,b,c,d,e;
}x[N];
int check(int i,int j,int k) {
    int sum=(x[j].a-x[i].a)*(x[k].a-x[i].a)+(x[j].b-x[i].b)*(x[k].b-x[i].b)+(x[j].c-x[i].c)*(x[k].c-x[i].c)
        +   (x[j].d-x[i].d)*(x[k].d-x[i].d)+(x[j].e-x[i].e)*(x[k].e-x[i].e);
    return sum;
}
int main() {
    int n;
    scanf("%d",&n);
    for(int i = 1; i <= n; ++i) {
        scanf("%d%d%d%d%d",&x[i].a,&x[i].b,&x[i].c,&x[i].d,&x[i].e);
    }
    for(int i=1;i<=n;++i) {
        int t= 0;
        if(vis[i]) continue;
        for(int j=1;j<=n;++j) {
            for(int k = 1; k <= n; ++k) {
                if(i == j || j ==k || i == k) continue;
                if(check(i,j,k) > 0) {
                    t = 1;
                    break;
                }
                else {
                    vis[j] = 1;
                    vis[k] = 1;
                }
            }
            if(t) break;
        }
        if(t) vis[i] = 1;
    }
    int sum= 0;
    for(int i = 1; i <= n; ++i) if(!vis[i]) ans[++sum] = i;
    cout<<sum<<endl;
    for(int i = 1; i <= sum; ++i) cout<<ans[i]<<endl;
    return 0;
}