zoj2688 Requirements Manhattan Distance

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zoj2688

Test instructions

Given n 5-D coordinates, find the maximum value of the Manhattan distance in these n coordinates.

Problem Solving Ideas:

Bare violence judgment Words time complexity is n^2, for n=10^5 of data amount, certainly will time out.

According to the two-dimensional coordinates of the Manhattan distance definition:

Dis (P1,P2) = ABS (X1-X2) + ABS (Y1-Y2);

If the absolute value is eliminated (i.e. the positive or negative of the enumeration symbol) There are four possibilities, namely the 2^k species (k=2)

Dis (P1,P2) = max{(x1+y1)-(X2+y2), (x1-y1)-(X2-y2), (-x1+y1)-(-x2+y2), (-x1-y1)-(-x2-y2)};

And in each case the x1,x2,x3 ... Same symbol, y1,y2,y3 .... Symbols are the same.

So. For 5-dimensional coordinates, we can enumerate all cases of 5 symbols (2^5) and save them.

Finally, the maximum distance can be compared with the same symbol. Time complexity is n*2^k (k=5)

Code:

#include <iostream> #include <cstdio> #include <cstring> #define MAXN 100050using namespace std;double        V[MAXN][32];d ouble a[6];void dfs (int x,int level,int b,double sum) {if (level==6) {v[x][b]=sum;    Return    } dfs (X,level+1,b<<1,sum+a[level]); DFS (x,level+1, (b<<1) +1,sum-a[level]);}    int main () {//Freopen ("In.txt", "R", stdin);    int n;    Double min1,max1;    Double ans;        while (~SCANF ("%d", &n) &&n) {ans=0.0;            for (int. i=1;i<=n;i++) {for (int j=1;j<=5;j++) scanf ("%lf", &a[j]);        DFS (i,1,0,0);            } for (int i=0;i<32;i++) {max1=min1=v[1][i];                for (int j=1;j<=n;j++) {if (Max1<v[j][i]) max1=v[j][i];            if (Min1>v[j][i]) min1=v[j][i];        } if (ans<max1-min1) ans=max1-min1;    } printf ("%.2lf\n", ans);} return 0;} 

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zoj2688 Requirements Manhattan Distance