【hihocoder 1628】K-Dimensional Foil（线性代数）

hihocoder 1627

The 2017 ACM-ICPC Asia Beijing Regional Contest 北京区域赛 B、K-Dimensional Foil

题意

给定N个点的前3维左边，和他们的欧几里得距离，求至少多少维，才能满足这个距离。

题解

施密特正交化可证明如果有解则存在下三角矩阵的解。距离平方和先减去前3维的距离平方和，这样就相当于去掉了3维。然后依次考虑每个点，看当前维度能不能满足答案，不能则加一维，再根据距离确定新加一维的值。

代码

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(i,l,r) for(int i=l,ed=r;i<ed;++i)
#define db(x) cout<< #x <<"="<<(x)<<endl
#define sqr(x) ((x)*(x))

typedef long long ll;
typedef long double dd;
const dd EPS=1e-10;
const int N=110;
int n,t;
dd a[N][N];//position of i_th point in j_th dimension
dd d[N][N];//remain distance between i_th and j_th point 
int num[N];//k_th dimension first appears on num[k]_th point
dd calc(int i,int j,int dim){//distance between j_th point and i_th point (dimension 0~dim)
	dd sum=0;
	rep(k,0,dim+1)
		sum+=sqr(a[j][k]-a[i][k]);
	return sum;
}
bool solve(){
	cin>>n;
	rep(i,0,n)
	rep(j,0,3)
		cin>>a[i][j];
	int flag=0;
	rep(i,0,n)
	rep(j,i+1,n){
		cin>>d[i][j];
		rep(k,0,3)d[i][j]-=sqr(a[i][k]-a[j][k]);
		if(d[i][j]<-EPS){
			flag=1;
		}
		d[j][i]=d[i][j];
	}
	if(flag)return 0;
	mem(a,0);
	mem(num,0);
	int k=0;
	rep(i,1,n){
		dd dis0=d[i][0];
		rep(j,0,k){
			if(a[num[j]][j]>EPS)
				a[i][j]=(calc(i,num[j],k)-calc(i,0,k)+d[i][0]-d[i][num[j]])/2./a[num[j]][j];
			dis0-=sqr(a[i][j]);
			if(dis0<-EPS)return 0;
		}
		if(dis0>EPS){
			a[num[k]=i][k]=sqrt(dis0);
			k++;
		}
		rep(j,0,i)
			if(fabs(calc(i,j,k)-d[i][j])>EPS)return 0;
	}
//	rep(i,0,n)
//	rep(j,0,k){
//		cout<<a[i][j]<<(" 
"[j==k-1]);
//	}
	cout<<k+3<<endl;
	return 1;
}
int main(){
	ios::sync_with_stdio(false);
	cin>>t;
	while(t--){
		if(!solve())cout<<"Goodbye World!"<<endl;
	}
	return 0;
}