POJ-2926 Requirements 最远曼哈顿距离

　　题目链接：http://poj.org/problem?id=2926

　　题意：求5维空间的点集中的最远曼哈顿距离。。

　　降维处理，推荐2009武森《浅谈信息学竞赛中的“0”和“1”》以及《论一类平面点对曼哈顿距离问题》。

 //STATUS:C++_AC_735MS_184KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 //const LL MOD=1000000007,STA=8000010;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e50;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 double x[],low[],hig[];
 int n;

 int main(){
  //   freopen("in.txt","r",stdin);
     int i,j,k;
     double ans,sum;
     while(~scanf("%d",&n))
     {
         for(i=;i<;i++){
             low[i]=OO;
             hig[i]=-OO;
         }
         for(i=;i<n;i++){
             scanf("%lf%lf%lf%lf%lf",&x[],&x[],&x[],&x[],&x[]);
             for(j=;j<;j++){
                 sum=;
                 for(k=;k<;k++){
                     sum+=(j&(<<k)?x[k]:-x[k]);
                 }
                 low[j]=Min(low[j],sum);
                 hig[j]=Max(hig[j],sum);
             }
         }
         ans=;
         for(i=;i<;i++){
             ans=Max(ans,hig[i]-low[i]);
         }

         printf("%.2lf\n",ans);
     }
     return ;
 }