You are given set of n points in 5-dimensional space. The points are labeled from 1 to n. No two points coincide.
We will call point a bad if there are different points b and c, not equal to a, from the given set such that angle between vectors 
The angle between vectors 
Given the list of points, print the indices of the good points in ascending order.
The first line of input contains a single integer n (1 ≤ n ≤ 103) — the number of points.
The next n lines of input contain five integers ai, bi, ci, di, ei (|ai|, |bi|, |ci|, |di|, |ei| ≤ 103) — the coordinates of the i-th point. All points are distinct.
First, print a single integer k — the number of good points.
Then, print k integers, each on their own line — the indices of the good points in ascending order.
6
0 0 0 0 0
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
1
1
3
0 0 1 2 0
0 0 9 2 0
0 0 5 9 0
0
In the first sample, the first point forms exactly a
In the second sample, along the cd plane, we can see the points look as follows:
We can see that all angles here are acute, so no points are good.
题意：
给你n个五维空间上的n个点。
让你求出“好”的点的数量。
“好”的点的定义是：
对于一个点i，不存在任何其他两个点，j,k， （三个点互不相同） 使得向量 i->j和向量 i - > k 是锐角。
思路：
判断两个向量的夹角是否是锐角可以通过向量的点积来判断。
五维空间的点积和二维三维的都一样，都是对应坐标相乘再相加。
然后每一次点积后的结果都可以得出这样的结论。
如图，如果点积后的结果小于等于零，那么证明∠J I K不是锐角。
那么就得出结论 j 和k都不是good点，因为对于j，有i和k是的所成的角是锐角。
k同理。
而如果点积结果大于0，那么证明 角jik是锐角，所以i不是good point
需要一个数组vis来维护哪些点是已经被确定不是good点的，以此来降低时间复杂度。
注意到枚举的时候，第三个变量k，设为j+1开始枚举，这样不会让i，j，k每一次重复计算而导致答案错误。
细节见代码：
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #include <vector> #include <iomanip> #define ALL(x) (x).begin(), (x).end() #define rt return #define dll(x) scanf("%I64d",&x) #define xll(x) printf("%I64d\n",x) #define sz(a) int(a.size()) #define all(a) a.begin(), a.end() #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), '\0', sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define eps 1e-6 #define gg(x) getInt(&x) #define db(x) cout<<"== [ "<<x<<" ] =="<<endl; using namespace std; typedef long long ll; ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} ll lcm(ll a,ll b){return a/gcd(a,b)*b;} ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;} inline void getInt(int* p); const int maxn=1010; const int inf=0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ int vis[maxn]; struct node { ll a,b,c,d,e; }arr[maxn]; int n; ll f(int i,int j,int k) { ll a1=arr[j].a-arr[i].a; ll a2=arr[k].a-arr[i].a; ll b1=arr[j].b-arr[i].b; ll b2=arr[k].b-arr[i].b; ll c1=arr[j].c-arr[i].c; ll c2=arr[k].c-arr[i].c; ll d1=arr[j].d-arr[i].d; ll d2=arr[k].d-arr[i].d; ll e1=arr[j].e-arr[i].e; ll e2=arr[k].e-arr[i].e; ll res=a1*a2+b1*b2+c1*c2+d1*d2+e1*e2; return res; } int main() { //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin); //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout); gbtb; cin>>n; repd(i,1,n) { cin>>arr[i].a>>arr[i].b>>arr[i].c>>arr[i].d>>arr[i].e; } repd(i,1,n) { if(i==2) { i=2; } if(vis[i]) continue; repd(j,1,n) { if(i==j) { continue; } repd(k,j+1,n) { if(k==j||k==i) { continue; } if(f(i,j,k)<=0) { vis[j]=vis[k]=1; break; }else { vis[i]=1; break; } } if(vis[i]) break; } } int ans=0; repd(i,1,n) { if(!vis[i]) { ans++; } } cout<<ans<<endl; repd(i,1,n) { if(!vis[i]) { cout<<i<<" "; } }cout<<endl; return 0; } inline void getInt(int* p) { char ch; do { ch = getchar(); } while (ch == ' ' || ch == '\n'); if (ch == '-') { *p = -(getchar() - '0'); while ((ch = getchar()) >= '0' && ch <= '9') { *p = *p * 10 - ch + '0'; } } else { *p = ch - '0'; while ((ch = getchar()) >= '0' && ch <= '9') { *p = *p * 10 + ch - '0'; } } }
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