POJ-2926 Requirements 最远曼哈顿距离

　　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 ;

 }