P1044 [NOIP2003 普及组] 栈

// Problem: P1044 [NOIP2003 普及组] 栈
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P1044
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// User: Pannnn

#include <bits/stdc++.h>

using namespace std;

template<class T>
void printVector(const T &a) {
    cout << "[ ";
    for (size_t i = 0; i < a.size(); ++i) {
        cout << a[i] << (i == a.size() - 1 ? " " : ", ");
    }
    cout << "]" << endl;
}

template<class T>
vector<vector<T>> matrix2(size_t n, size_t m, T init) {
    return vector<vector<T>>(n, vector<T>(m, init));
}

/*
void dfs(vector<vector<int>> &res, stack<int> st, vector<int> tmp, int cur, int n) {
    if (cur > n) {
        while (!st.empty()) {
            int t = st.top();
            st.pop();
            tmp.push_back(t);
        }
        res.push_back(tmp);
        return;
    }
    
    st.push(cur);
    dfs(res, st, tmp, cur + 1, n);
    st.pop();
    
    if (!st.empty()) {
        int t = st.top();
        st.pop();
        tmp.push_back(t);
        dfs(res, st, tmp, cur, n);
    }
}
*/
// 模拟 超时
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    /*
    int n;
    cin >> n;
    vector<vector<int>> res;
    stack<int> st;
    vector<int> tmp;
    
    dfs(res, st, tmp, 1, n);
    cout << res.size() << endl;
    */
    
    /*
        定义状态：
        f[i][j] i表示栈内数字的个数，j表示未进栈数字的个数，f为当前状态下有几种情况
        栈里的数字有两种选择：出栈和不出栈
        若出栈，栈里数字个数减一
        如果不出栈，未进栈的数字要进来一个压栈，栈内个数加一，未进栈个数减一
        f[i][j] = f[i - 1][j] + f[i + 1][j - 1];
        边界：
        当栈内没有数字时，只能进栈，且此操作后的出栈情况取决于f[i + 1][j - 1];
        当栈外没有数字时，只能出栈：f[i][0] = 1;
    */
    int n;
    cin >> n;
    auto info = matrix2<int>(n + 2, n + 2, 0);
    for (int i = 0; i <= n; ++i) {
        info[i][0] = 1;
    }
    
    for (int j = 1; j <= n; ++j) {
        for (int i = 0; i <= n; ++i) {
            if (i >= 1) {
                info[i][j] = info[i - 1][j] + info[i + 1][j - 1];
            } else {
                info[i][j] = info[i + 1][j - 1];
            }
        }
    }
    cout << info[0][n] << endl;
    return 0;
}