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》。
 1 //STATUS:C++_AC_735MS_184KB
 2 #include <functional>
 3 #include <algorithm>
 4 #include <iostream>
 5 //#include <ext/rope>
 6 #include <fstream>
 7 #include <sstream>
 8 #include <iomanip>
 9 #include <numeric>
10 #include <cstring>
11 #include <cassert>
12 #include <cstdio>
13 #include <string>
14 #include <vector>
15 #include <bitset>
16 #include <queue>
17 #include <stack>
18 #include <cmath>
19 #include <ctime>
20 #include <list>
21 #include <set>
22 #include <map>
23 using namespace std;
24 //#pragma comment(linker,"/STACK:102400000,102400000")
25 //using namespace __gnu_cxx;
26 //define
27 #define pii pair<int,int>
28 #define mem(a,b) memset(a,b,sizeof(a))
29 #define lson l,mid,rt<<1
30 #define rson mid+1,r,rt<<1|1
31 #define PI acos(-1.0)
32 //typedef
33 typedef __int64 LL;
34 typedef unsigned __int64 ULL;
35 //const
36 const int N=50010;
37 const int INF=0x3f3f3f3f;
38 //const LL MOD=1000000007,STA=8000010;
39 const LL LNF=1LL<<55;
40 const double EPS=1e-9;
41 const double OO=1e50;
42 const int dx[4]={-1,0,1,0};
43 const int dy[4]={0,1,0,-1};
44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
45 //Daily Use ...
46 inline int sign(double x){return (x>EPS)-(x<-EPS);}
47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
50 template<class T> inline T Min(T a,T b){return a<b?a:b;}
51 template<class T> inline T Max(T a,T b){return a>b?a:b;}
52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
56 //End
57 
58 double x[5],low[32],hig[32];
59 int n;
60 
61 int main(){
62  //   freopen("in.txt","r",stdin);
63     int i,j,k;
64     double ans,sum;
65     while(~scanf("%d",&n))
66     {
67         for(i=0;i<32;i++){
68             low[i]=OO;
69             hig[i]=-OO;
70         }
71         for(i=0;i<n;i++){
72             scanf("%lf%lf%lf%lf%lf",&x[0],&x[1],&x[2],&x[3],&x[4]);
73             for(j=0;j<32;j++){
74                 sum=0;
75                 for(k=0;k<5;k++){
76                     sum+=(j&(1<<k)?x[k]:-x[k]);
77                 }
78                 low[j]=Min(low[j],sum);
79                 hig[j]=Max(hig[j],sum);
80             }
81         }
82         ans=0;
83         for(i=0;i<32;i++){
84             ans=Max(ans,hig[i]-low[i]);
85         }
86 
87         printf("%.2lf
",ans);
88     }
89     return 0;
90 }