HDU 4666 Hyperspace

Hyperspace

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 136    Accepted Submission(s): 66

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 20 208 4030 371 -1801 20 1069 -1920 418 -5251 51 10 2754 6350 -2491 9610 2954 -2516

 

Sample Output

074601456145614560251255718922

 

Source

2013 Multi-University Training Contest 7

 

Recommend

zhuyuanchen520

 

题意： 有Q个操作。 没次操作会增加一个点， 或者删除一个点。 每次输出点集的最大曼哈顿距离。

思路： STL应用

不知道什么原理，师兄说的。

一维就是  Max (x) - Min(x)

就是对于 二维的 x - y   和 x + y 做两个集合。 答案肯定会在  Max( x - y) -  Min( x - y)  或者 是  Max(x + y)  -  Min(x + y)

而三维就是 Max(x + y + z) - Min(x + y + z)  或者是 Max(x + y - z)  - Min(x + y - z) 略。

就是  a +- b +- c +- d +- e

2 ^ (k - 1)个集合里。

#include <cstdio>#include <cstring>#include <algorithm>#include <set>#include <map>using namespace std;//const int V = 60000 + 50;const int inf = 0x7fffffff;int n, k;int main() {    int i, j, q;    while(~scanf("%d%d", &n, &k)) {        multiset<int> s[20];        map<int, int> m[20];        for(i = 1; i <= n; ++i) {            int op;            scanf("%d", &op);            if(op == 0) {                int po[10];                for(j = 0; j < k; ++j)                    scanf("%d", &po[j]);                for(j = 0; j < (1 << (k - 1)); ++j) {                    int sum = po[0];                    for(q = 0; q < k - 1; ++q) {                        if(j & (1 << q))                            sum -= po[q + 1];                        else                            sum += po[q + 1];                    }                    s[j].insert(sum);                    m[j][i] = sum;                }            }            else {                int temp;                scanf("%d", &temp);                for(j = 0; j < (1 << (k - 1)); ++j) {                    if(m[j].size() > 0) {                        s[j].erase(s[j].find(m[j][temp]));                        m[j].erase(temp);                    }                }            }            int Max = 0;            for(j = 0; j < (1 << (k - 1)); ++j) {                if(s[j].size() > 0) {                    multiset<int>::iterator fi = s[j].begin();                    multiset<int>::iterator se = s[j].end();                    se--;                    Max = max(Max, *se - *fi);                }            }            printf("%d\n", Max);        }    }}