poj 2926 Requirements 【5维点集最远曼哈顿距离】

最新推荐文章于 2018-12-27 04:37:00 发布

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本文探讨了一名大学生为了提高被研究生院录取的机会而进行的独立研究，重点关注了他为完成本科教育所需满足的五项独特要求。通过比较不同大学的课程要求，学生发现这些要求在整体上具有相似性，并提出了一种衡量不同大学课程要求差异的方法。

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Requirements

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 4090		Accepted: 1411

Description

An undergraduate student, realizing that he needs to do research to improve his chances of being accepted to graduate school, decided that it is now time to do some independent research. Of course, he has decided to do research in the most important domain: the requirements he must fulfill to graduate from his undergraduate university. First, he discovered (to his surprise) that he has to fulfill 5 distinct requirements: the general institute requirement, the writing requirement, the science requirement, the foreign-language requirement, and the field-of-specialization requirement. Formally, a requirement is a fixed number of classes that he has to take during his undergraduate years. Thus, for example, the foreign language requirement specifies that the student has to take 4 classes to fulfill this requirement: French I, French II, French III, and French IV. Having analyzed the immense multitude of the classes that need to be taken to fulfill the different requirements, our student became a little depressed about his undergraduate university: there are so many classes to take…

Dejected, the student began studying the requirements of other universities that he might have chosen after high school. He found that, in fact, other universities had exactly the same 5 requirements as his own university. The only difference was that different universities had different number of classes to be satisfied in each of the five requirement.

Still, it appeared that universities have pretty similar requirements (all of them require a lot of classes), so he hypothesized that no two universities are very dissimilar in their requirements. He defined the dissimilarity of two universities X and Y as |x₁ − y₁| + |x₂ − y₂| + |x₃ − y₃| + |x₄ − y₄| + |x₅ − y₅|, where an x_i (y_i) is the number of classes in the requirement i of university X (Y) multiplied by an appropriate factor that measures hardness of the corresponding requirement at the corresponding university.

Input

The first line of the input file contains an integer N (1 ≤ N ≤ 100 000), the number of considered universities. The following N lines each describe the requirements of a university. A university X is described by the five non-negative real numbers x₁ x₂ x₃ x₄ x₅.

Output

On a single line, print the dissimilarity value of the two most dissimilar universities. Your answer should be rounded to exactly two decimal places.

Sample Input

3
2 5 6 2 1.5
1.2 3 2 5 4
7 5 3 2 5

Sample Output

12.80

题意：给顶n个5维空间的点集，问你两点间最远的曼哈顿距离。

思路：用二进制枚举5个坐标的正负情况，对每一种情况找出最优值，最终的结果就是所有最优值里面的最大值。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <string>
#define INF (2000000000+10)
#define eps 1e-8
#define MAXN (300000+10)
#define MAXM (600000+10)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while((a)--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
#pragma comment(linker, "/STACK:102400000,102400000")
#define fi first
#define se second
using namespace std;
typedef pair<int, int> pii;
double a[MAXN][5];
int main()
{
    int n;
    while(Ri(n) != EOF)
    {
        for(int i = 0; i < n; i++)
            for(int j = 0; j < 5; j++)
                Rf(a[i][j]);
        double ans = 0, mi, mx, temp;
        for (int s = 0; s < (1<<5); s++)
        {
            mi = INF, mx = -INF;
            for(int i = 0; i < n; i++)
            {
                temp = 0;
                for(int j = 0; j < 5; j++)
                {
                    if ((1<<j) & s) temp += a[i][j];
                    else temp -= a[i][j];
                }
                mi = min(mi, temp);
                mx = max(mx, temp);
            }
            ans = max(ans, mx-mi);
        }
        Pf(ans);
    }
    return 0;
}