poj 2926 Requirements (曼哈顿距离)

题目链接

题目描述:

Hyperspace

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)

Total Submission(s): 1262 Accepted Submission(s): 611

Problem Description

The great Mr.Smith has invented a hyperspace particle generator. The device is very powerful. The device can generate a hyperspace. In the hyperspace, particle may appear and disappear randomly. At the same time a great amount of energy was generated.

However, the device is in test phase, often in a unstable state. Mr.Smith worried that it may cause an explosion while testing it. The energy of the device is related to the maximum manhattan distance among particle.

Particles may appear and disappear any time. Mr.Smith wants to know the maxmium manhattan distance among particles when particle appears or disappears.

Input

The input contains several test cases, terminated by EOF.

In each case: In the first line, there are two integer q(number of particle appear and disappear event, ≤60000) and k(dimensions of the hyperspace that the hyperspace the device generated, ≤5). Then follows q lines. In each line, the first integer ‘od’ represents the event: od = 0 means this is an appear

event. Then follows k integer(with absolute value less then 4 × 107). od = 1 means this is an disappear event. Follows a integer p represents the disappeared particle appeared in the pth event.

Output

Each test case should contains q lines. Each line contains a integer represents the maximum manhattan distance among paticles.

Sample Input

10 2

0 208 403

0 371 -180

1 2

0 1069 -192

0 418 -525

1 5

1 1

0 2754 635

0 -2491 961

0 2954 -2516

Sample Output

0

746

0

1456

1456

1456

0

2512

5571

8922

题意：

给出n个5维点，计算两点的曼哈顿距离的最大值。

题解：

传送门~

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn=1e5+5;
const double inf=0x3f3f3f3f;
double a[maxn][6];
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=1;i<=n;i++)
for(int j=1;j<=5;j++)
scanf("%lf",&a[i][j]);
double ans=0;
for(int i=0;i<(1<<5);i++)
{
double mx=-inf,mi=inf;
for(int j=1;j<=n;j++)
{
double p=0;
for(int k=0;k<5;k++)
{
if(i&(1<<k)) p+=a[j][k+1];
else p-=a[j][k+1];
}
mx=max(mx,p);
mi=min(mi,p);
}
ans=max(ans,mx-mi);
}
printf("%.2lf\n",ans);
}
}