1057 Stack

Stack is one of the most fundamental data structures, which is based on the principle of Last In First Out (LIFO). The basic operations include Push (inserting an element onto the top position) and Pop (deleting the top element). Now you are supposed to implement a stack with an extra operation: PeekMedian -- return the median value of all the elements in the stack. With N elements, the median value is defined to be the (-th smallest element if N is even, or (-th if N is odd.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤). Then N lines follow, each contains a command in one of the following 3 formats:

Push key
Pop
PeekMedian

 

where key is a positive integer no more than 1.

Output Specification:

For each Push command, insert key into the stack and output nothing. For each Pop or PeekMedian command, print in a line the corresponding returned value. If the command is invalid, print Invalid instead.

Sample Input:

17
Pop
PeekMedian
Push 3
PeekMedian
Push 2
PeekMedian
Push 1
PeekMedian
Pop
Pop
Push 5
Push 4
PeekMedian
Pop
Pop
Pop
Pop

 

Sample Output:

Invalid
Invalid
3
2
2
1
2
4
4
5
3
Invalid

题意：

　　模拟一个stack，并且新增一个返回中值的功能。

思路：

　　如果每次查询中值都sort一次的话，只能通过两组测试点。看了别人的blog之后发现这道题有很多种做法，其中较好理解的一种就是用二维桶排序的方法来做这道题。大体的思路是这样的：首先划分出100个大桶，每一个大桶中存放1000个小桶。用数组下标来表示栈中的数字，数组的值来表示出现的次数。另外还要用一个栈来模拟push和pop，桶排序用来查找中值。先找出目标数字出现在那个大桶中，然后再在大桶中找出在那个小桶中。

Code：

#include <bits/stdc++.h>

using namespace std;

int main() {
    int n;
    cin >> n;
    getchar();
    string str;
    stack<int> stk;
    int prt = -1;
    int bucket_count[100] = {0};
    int bucket[100][1000] = {0};
    for (int i = 0; i < n; ++i) {
        getline(cin, str);
        if (str[1] == 'u') {
            int num = stoi(str.substr(5));
            bucket_count[num / 1000]++;
            bucket[num / 1000][num % 1000]++;
            stk.push(num);
        } else if (str[1] == 'o') {
            if (stk.empty())
                cout << "Invalid" << endl;
            else {
                int num = stk.top();
                cout << num << endl;
                bucket_count[num / 1000]--;
                bucket[num / 1000][num % 1000]--;
                stk.pop();
            }
        } else {
            if (stk.empty()) {
                cout << "Invalid" << endl;
            } else {
                int target = (stk.size() + 1) / 2;
                int count = 0;
                for (int i = 0; i < 100; ++i) {
                    if (count + bucket_count[i] >= target) {
                        for (int j = 0; j < 1000; ++j) {
                            if (count + bucket[i][j] >= target) {
                                cout << i * 1000 + j << endl;
                                break;
                            }
                            count += bucket[i][j];
                        }
                        break;
                    }
                    count += bucket_count[i];
                }
            }
        }
    }

    return 0;
}