题解 | 吐泡泡

#include <iostream>
#include <stack>
#include <string>
#include <vector>
using namespace std;

int main() {
    int T;
    cin >> T;

    while (T--) {
        string zifuchuan;
        cin >> zifuchuan;

        if (zifuchuan.empty()) {
            cout << endl;
            continue;
        }

        stack<char> zhan;
        zhan.push(zifuchuan[0]);

        for (int i = 1; i < zifuchuan.size(); i++) {
            char current = zifuchuan[i];
            
            if (zhan.empty()) {
                zhan.push(current);
                continue;
            }
            
            char top = zhan.top();
            
            if (current == 'O' && top == 'O') {
                // 两个大泡泡爆炸
                zhan.pop();
            } else if (current == 'o' && top == 'o') {
                // 两个小泡泡融合成大泡泡
                zhan.pop();
                zhan.push('O');
                
                // 检查新产生的大泡泡是否引发连锁反应
                if (zhan.size() >= 2) {
                    char new_top = zhan.top();  // 新产生的'O'
                    zhan.pop();
                    char next_top = zhan.top(); // 原来的栈顶
                    
                    if (new_top == 'O' && next_top == 'O') {
                        // 新产生的大泡泡和原来的大泡泡爆炸
                        zhan.pop();
                    } else {
                        zhan.push(new_top);
                    }
                }
            } else {
                zhan.push(current);
            }
        }

        // 正确地从栈中提取结果
        vector<char> result;
        while (!zhan.empty()) {
            result.push_back(zhan.top());
            zhan.pop();
        }

        // 反转结果（因为栈是后进先出）
        for (int i = result.size() - 1; i >= 0; i--) {
            cout << result[i];
        }
        cout << endl;
    }
    return 0;
}