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]<<(" \n"[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;
}