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;
}