HDU 5818 Joint Stacks 栈模拟

最新推荐文章于 2020-08-15 10:56:42 发布

Southan97

最新推荐文章于 2020-08-15 10:56:42 发布

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分类专栏： Implementation 文章标签：栈模拟

本文链接：https://blog.youkuaiyun.com/Mrx_Nh/article/details/71154045

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本文介绍了一种数据结构——可合并栈（mergeable stack），并提供了详细的实现方案。该栈支持三种基本操作：push、pop 和 merge，并通过样例解释了操作流程。文章还提供了一种高效实现方式，确保整体时间复杂度为 O(N)。

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Joint Stacks

Problem Description

A stack is a data structure in which all insertions and deletions of entries are made at one end, called the "top" of the stack. The last entry which is inserted is the first one that will be removed. In another word, the operations perform in a Last-In-First-Out (LIFO) manner.
A mergeable stack is a stack with "merge" operation. There are three kinds of operation as follows:

- push A x: insert x into stack A
- pop A: remove the top element of stack A
- merge A B: merge stack A and B

After an operation "merge A B", stack A will obtain all elements that A and B contained before, and B will become empty. The elements in the new stack are rearranged according to the time when they were pushed, just like repeating their "push" operations in one stack. See the sample input/output for further explanation.
Given two mergeable stacks A and B, implement operations mentioned above.

Input

There are multiple test cases. For each case, the first line contains an integer

N(0<N≤105), indicating the number of operations. The next N lines, each contain an instruction "push", "pop" or "merge". The elements of stacks are 32-bit integers. Both A and B are empty initially, and it is guaranteed that "pop" operation would not be performed to an empty stack. N = 0 indicates the end of input.

Output

For each case, print a line "Case #t:", where t is the case number (starting from 1). For each "pop" operation, output the element that is popped, in a single line.

Sample Input

4
push A 1
push A 2
pop A
pop A
9
push A 0
push A 1
push B 3
pop A
push A 2
merge A B
pop A
pop A
pop A
9
push A 0
push A 1
push B 3
pop A
push A 2
merge B A
pop B
pop B
pop B 
0

Sample Output

Case #1:
2
1
Case #2:
1
2
3
0
Case #3:
1
2
3
0

Joint Stacks

比较简单巧妙的一个做法是引入一个新的栈C，每次合并的时候就把A和B合并到C上，然后把A和B都清空. push还是按正常做，pop注意当遇到要pop的栈为空时，因为题目保证不会对空栈进行pop操作，所以这时应直接改为对C栈进行pop操作. 这样做因为保证每个元素最多只在一次合并中被处理到，pop和push操作当然也是每个元素只做一次，所以总复杂度是O(N)的. 另一种做法是用链表来直接模拟，复杂度也是O(N)，但代码量稍大一些.

A, B入栈的时候加个时间戳，方便合并。

#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
const int MAXN = 100000 + 7;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int n;
    int cas = 1;
    while (cin >> n) {
        if (!n) break;
        cout << "Case #" << cas++ << ":" << endl;
        stack<pair<int, int> > a;
        stack<pair<int, int> > b;
        stack<int> c;
        string op;
        char cc;
        int num;
        int t = 0;
        for (int i = 0; i < n; ++i) {
            cin >> op;
            if (op == "push") {
                cin >> cc >> num;
                if (cc == 'A') {
                    a.push(make_pair(num, t++));
                } else {
                    b.push(make_pair(num, t++));
                }
            } else if (op == "pop") {
                cin >> cc;
                if (cc == 'A') {
                    if (!a.empty()) {
                        cout << a.top().first << endl;
                        a.pop();
                    } else {
                        cout << c.top() << endl;
                        c.pop();
                    }
                } else {
                    if (!b.empty()) {
                        cout << b.top().first << endl;
                        b.pop();
                    } else {
                        cout << c.top() << endl;
                        c.pop();
                    }
                }
            } else {
                char s1, s2;
                cin >> s1 >> s2;
                if (s1 == s2) continue;
                else {
                    stack<int> tmp;
                    while (!a.empty() || !b.empty()) {
                        if (a.empty()) {
                            while (!b.empty()) {
                                tmp.push(b.top().first);
                                b.pop();
                            }
                        }
                        if (b.empty()) {
                            while (!a.empty()) {
                                tmp.push(a.top().first);
                                a.pop();
                            }
                        }
                        if(!a.empty() && !b.empty()) {
                            if (a.top().second < b.top().second) {
                                tmp.push(b.top().first);
                                b.pop();
                            } else {
                                tmp.push(a.top().first);
                                a.pop();
                            }
                        }
                    }
                    while (!tmp.empty()) {
                        c.push((int &&) tmp.top());
                        tmp.pop();
                    }
                }
            }
        }
    }

return 0;
}