hdu 4666 Hyperspace （曼哈顿距离+set ）

Hyperspace

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

Total Submission(s): 931    Accepted Submission(s): 441

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 × 10 ⁷). 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
   
   
   
   

 

Source

2013 Multi-University Training Contest 7

 

思路：

先求曼哈顿距离（复杂度:O(n*k*2^k)   k-维数 ），例如二维的曼哈顿距离，有两点(x1,y1),(x2,y2)，那么它们的曼哈顿距离就是 dist=|x1-x2|+|y1-y2|,去绝对值得

dist=max（ x2+y2-(x1+y1) , (x2-y2)-(x1-y1) , (-x2+y2)-(-x1+y1) , (-x2-y2)-(-x1-y1) ）;

可以看出每个式子中的两点的符号都是一样的，那么问题就简化为

找到每个符号状态下的最大最小值相减取最大值就够了。

扩展到多维是一样的，求曼哈顿距离解决了，然后用二进制来枚举状态，用multiset来维护插入删除就够了。

注意：

multiset中对值删除的话是删除所有的这些值，所以要找到一个指针后再删除。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 35
#define MAXN 60005
#define mod 100000007
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

int n,m,w,ans;
bool vis[MAXN];
int x[MAXN][5];
multiset<int>dist[maxn];
multiset<int>::iterator it;

int main()
{
    int i,j,k,t,op,d;
    while(~scanf("%d%d",&n,&w))
    {
        m=1<<w;
        for(i=0; i<m; i++)
        {
            dist[i].clear();
        }
        memset(vis,0,sizeof(vis));
        for(i=1; i<=n; i++)
        {
            scanf("%d",&op);
            if(!op)
            {
                for(j=0; j<w; j++)
                {
                    scanf("%d",&x[i][j]);
                }
                for(j=0; j<m; j++)
                {
                    d=0;
                    for(k=0; k<w; k++)
                    {
                        if(j&(1<<k)) d+=x[i][k];
                        else d-=x[i][k];
                    }
                    dist[j].insert(d);
                }
                ans=0;
                for(j=0; j<m; j++)
                {
                    if(!dist[j].empty())
                    {
                        it=dist[j].end();
                        it--;
                        ans=max(ans,*it-*dist[j].begin());
                    }
                }
            }
            else
            {
                scanf("%d",&t);
                if(!vis[t])
                {
                    vis[t]=1;
                    for(j=0; j<m; j++)
                    {
                        d=0;
                        for(k=0; k<w; k++)
                        {
                            if(j&(1<<k)) d+=x[t][k];
                            else d-=x[t][k];
                        }
                        it=dist[j].find(d);
                        dist[j].erase(it);
                    }
                }
                ans=0;
                for(j=0; j<m; j++)
                {
                    if(!dist[j].empty())
                    {
                        it=dist[j].end();
                        it--;
                        ans=max(ans,*it-*dist[j].begin());
                    }
                }
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}