pku 2926 Requirements 多维曼哈顿距离

题意：

给你5维的n个点的坐标，求其中两点之间曼哈顿距离最大。

思路：
多维曼哈顿距离的模板题目

|x1-x'1|+|x2-x'2|+... ...+|xm-x'm| 去掉绝对值后x,x'分别都有2^m种状态，枚举每种状态下的最大值和最下值，然后找出差值最大的即可。

#include <iostream>

#include <cstdio>

#include <cmath>

#include <vector>

#include <cstring>

#include <algorithm>

#include <string>

#include <set>

#include <functional>

#include <numeric>

#include <sstream>

#include <stack>

#include <map>

#include <queue>



#define CL(arr, val) memset(arr, val, sizeof(arr))



#define lc l,m,rt<<1

#define rc m + 1,r,rt<<1|1

#define pi acos(-1.0)

#define ll long long

#define L(x)    (x) << 1

#define R(x)    (x) << 1 | 1

#define MID(l, r)   (l + r) >> 1

#define Min(x, y)   (x) < (y) ? (x) : (y)

#define Max(x, y)   (x) < (y) ? (y) : (x)

#define E(x)        (1 << (x))

#define iabs(x)     (x) < 0 ? -(x) : (x)

#define OUT(x)  printf("%I64d\n", x)

#define lowbit(x)   (x)&(-x)

#define Read()  freopen("data.in", "r", stdin)

#define Write() freopen("d.out", "w", stdout)





#define M 100007

#define N 100007



using namespace std;





const int inf = 0x7f7f7f7f;

const int mod = 1000000007;



struct node

{

    double x[6];

}pt[N];



double maxx[1<<5],minx[1<<5];



int n,m;//n表示点数，m表示维数



void solve(int n,int dem)

{

    int msk = 1<<dem;

    int i,j,k;

    for (i = 0; i < msk; ++i)

    {

        maxx[i] = -inf;

        minx[i] = inf;

    }

    for (i = 0; i < n; ++i)

    {

        for (j = 0; j < msk; ++j)

        {

            double s = 0;

            int tmsk = j;

            for (k = 0; k < dem; ++k)

            {

                if (tmsk&1) s += pt[i].x[k];

                else s -= pt[i].x[k];

                tmsk >>= 1;

            }

            if (maxx[j] < s) maxx[j] = s;

            if (minx[j] > s) minx[j] = s;

        }

    }

    double ans = 0;

    for (i = 0; i < msk; ++i)

    {

        ans = max(ans,maxx[i] - minx[i]);

    }

    printf("%.2f\n",ans);

}

int main()

{

    int i,j;

    m = 5;

    while (~scanf("%d",&n))

    {



        for (i = 0; i < n; ++i)

        {

            for (j = 0; j < m; ++j)

            {

                scanf("%lf",&pt[i].x[j]);

            }

        }

        solve(n,m);

    }

    return 0;

}

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