Game of Stacks_信奥算法NOI/NOI+/CTSC-ACGO题库

题目描述

You have $n$ stacks $r_1,r_2,\ldots,r_n$ . Each stack contains some positive integers ranging from $1$ to $n$ .

Define the following functions:

function init(pos):    stacks := an array that contains n stacks r[1], r[2], ..., r[n]    return get(stacks, pos)function get(stacks, pos):    if stacks[pos] is empty:        return pos    else:        new_pos := the top element of stacks[pos]        pop the top element of stacks[pos]        return get(stacks, new_pos)

You want to know the values returned by $\texttt{init(1)}, \texttt{init(2)}, \ldots, \texttt{init(n)}$ .

Note that, during these calls, the stacks $r_1,r_2,\ldots,r_n$ don't change, so the calls $\texttt{init(1)}, \texttt{init(2)}, \ldots, \texttt{init(n)}$ are independent.

你有 $n$ 个栈 $r_1, r_2, \ldots, r_n$ 。每个栈中包含若干个取值在 $1$ 到 $n$ 之间的正整数。

定义如下函数：

function init(pos):    stacks := 一个包含 $n$ 个栈 $r[1], r[2], \ldots, r[n]$ 的数组    return get(stacks, pos)function get(stacks, pos):    if stacks[pos] 为空:        return pos    else:        new_pos := stacks[pos] 的栈顶元素        将 stacks[pos] 的栈顶元素弹出        return get(stacks, new_pos)

你需要求出 $\texttt{init(1)}, \texttt{init(2)}, \ldots, \texttt{init(n)}$ 各自返回的值。

注意：在这些调用过程中，栈 $r_1, r_2, \ldots, r_n$ 均不发生改变，因此调用 $\texttt{init(1)}, \texttt{init(2)}, \ldots, \texttt{init(n)}$ 是相互独立的。

输入格式

The first line of the input contains one integer $n$ ( $1\le n\le 10^5$ ) — the length of the array $r$ .

Each of the following $n$ lines contains several integers. The first integer $k_i$ ( $0\le k_i\le 10^5$ ) represents the number of elements in the $i$ -th stack, and the following $k_i$ positive integers $c_{i,1},c_{i,2},\ldots,c_{i,k_i}$ ( $1\le c_{i,j}\le n$ ) represent the elements in the $i$ -th stack. $c_{i,1}$ is the bottom element.

In each test, $\sum k_i\le 10^6$ .

输入的第一行包含一个整数 $n$ （ $1\le n\le 10^5$ ）—— 表示数组 $r$ 的长度。

接下来的 $n$ 行，每行包含若干个整数。第一个整数 $k_i$ （ $0\le k_i\le 10^5$ ）表示第 $i$ 个栈中元素的个数；随后的 $k_i$ 个正整数 $c_{i,1},c_{i,2},\ldots,c_{i,k_i}$ （ $1\le c_{i,j}\le n$ ）表示第 $i$ 个栈中的元素，其中 $c_{i,1}$ 为栈底元素。

在每个测试用例中，满足 $\sum k_i\le 10^6$ 。

输出格式

You need to output $n$ values, the $i$ -th of which is the value returned by $\texttt{init(i)}$ .

你需要输出 $n$ 个值，其中第 $i$ 个值为 $\texttt{init(i)}$ 返回的值。

输入输出样例

输入#1
```
3
3 1 2 2
3 3 1 2
3 1 2 1
```
输出#1
```
1 2 2
```

输入#2

5
5 1 2 4 3 4
6 1 2 5 3 3 4
6 1 1 4 4 4 2
9 3 1 4 2 3 5 5 1 2
4 4 4 1 3

输出#2

1 1 1 1 1

说明/提示

In the first example:

When you call $\texttt{init(1)}$ $init(1)$ , set $\texttt{stacks := [[1,2,2],[3,1,2],[1,2,1]]}$ $stacks := [[1,2,2],[3,1,2],[1,2,1]]$ , and then call $\texttt{get(stacks, 1)}$ $get(stacks, 1)$ .
- $\texttt{stacks[1]}$ is not empty, set \texttt{new_pos := 2}, and pop the top element of $\texttt{stacks[1]}$ , which makes $\texttt{stacks}$ become $[[1,2],[3,1,2],[1,2,1]]$ , and then call $\texttt{get(stacks, 2)}$ .
- $\texttt{stacks[2]}$ is not empty, set \texttt{new_pos := 2}, and pop the top element of $\texttt{stacks[2]}$ , which makes $\texttt{stacks}$ become $[[1,2],[3,1],[1,2,1]]$ , and then call $\texttt{get(stacks, 2)}$ .
- $\texttt{stacks[2]}$ is not empty, set \texttt{new_pos := 1}, and pop the top element of $\texttt{stacks[2]}$ , which makes $\texttt{stacks}$ become $[[1,2],[3],[1,2,1]]$ , and then call $\texttt{get(stacks, 1)}$ .
- $\texttt{stacks[1]}$ is not empty, set \texttt{new_pos := 2}, and pop the top element of $\texttt{stacks[1]}$ , which makes $\texttt{stacks}$ become $[[1],[3],[1,2,1]]$ , and then call $\texttt{get(stacks, 2)}$ .
- $\texttt{stacks[2]}$ is not empty, set \texttt{new_pos := 3}, and pop the top element of $\texttt{stacks[2]}$ , which makes $\texttt{stacks}$ become $[[1],[],[1,2,1]]$ , and then call $\texttt{get(stacks, 3)}$ .
- $\texttt{stacks[3]}$ is not empty, set \texttt{new_pos := 1}, and pop the top element of $\texttt{stacks[3]}$ , which makes $\texttt{stacks}$ become $[[1],[],[1,2]]$ , and then call $\texttt{get(stacks, 1)}$ .
- $\texttt{stacks[1]}$ is not empty, set \texttt{new_pos := 1}, and pop the top element of $\texttt{stacks[1]}$ , which makes $\texttt{stacks}$ become $[[],[],[1,2]]$ , and then call $\texttt{get(stacks, 1)}$ .
- $\texttt{stacks[1]}$ is empty, return $1$ .
When you call $\texttt{init(2)}$ $init(2)$ , set $\texttt{stacks := [[1,2,2],[3,1,2],[1,2,1]]}$ $stacks := [[1,2,2],[3,1,2],[1,2,1]]$ , and then call $\texttt{get(stacks, 2)}$ $get(stacks, 2)$ .
- $\texttt{stacks[2]}$ is not empty, set \texttt{new_pos := 2}, and pop the top element of $\texttt{stacks[2]}$ , which makes $\texttt{stacks}$ become $[[1,2,2],[3,1],[1,2,1]]$ , and then call $\texttt{get(stacks, 2)}$ .
- $\texttt{stacks[2]}$ is not empty, set \texttt{new_pos := 1}, and pop the top element of $\texttt{stacks[2]}$ , which makes $\texttt{stacks}$ become $[[1,2,2],[3],[1,2,1]]$ , and then call $\texttt{get(stacks, 1)}$ .
- $\texttt{stacks[1]}$ is not empty, set \texttt{new_pos := 2}, and pop the top element of $\texttt{stacks[1]}$ , which makes $\texttt{stacks}$ become $[[1,2],[3],[1,2,1]]$ , and then call $\texttt{get(stacks, 2)}$ .
- $\texttt{stacks[2]}$ is not empty, set \texttt{new_pos := 3}, and pop the top element of $\texttt{stacks[2]}$ , which makes $\texttt{stacks}$ become $[[1,2],[],[1,2,1]]$ , and then call $\texttt{get(stacks, 3)}$ .
- $\texttt{stacks[3]}$ is not empty, set \texttt{new_pos := 1}, and pop the top element of $\texttt{stacks[3]}$ , which makes $\texttt{stacks}$ become $[[1,2],[],[1,2]]$ , and then call $\texttt{get(stacks, 1)}$ .
- $\texttt{stacks[1]}$ is not empty, set \texttt{new_pos := 2}, and pop the top element of $\texttt{stacks[1]}$ , which makes $\texttt{stacks}$ become $[[1],[],[1,2]]$ , and then call $\texttt{get(stacks, 2)}$ .
- $\texttt{stacks[2]}$ is empty, return $2$ .
When you call $\texttt{init(3)}$ $init(3)$ , set $\texttt{stacks := [[1,2,2],[3,1,2],[1,2,1]]}$ $stacks := [[1,2,2],[3,1,2],[1,2,1]]$ , and then call $\texttt{get(stacks, 3)}$ $get(stacks, 3)$ .
- $\texttt{stacks[3]}$ is not empty, set \texttt{new_pos := 1}, and pop the top element of $\texttt{stacks[3]}$ , which makes $\texttt{stacks}$ become $[[1,2,2],[3,1,2],[1,2]]$ , and then call $\texttt{get(stacks, 1)}$ .
- $\texttt{stacks[1]}$ is not empty, set \texttt{new_pos := 2}, and pop the top element of $\texttt{stacks[1]}$ , which makes $\texttt{stacks}$ become $[[1,2],[3,1,2],[1,2]]$ , and then call $\texttt{get(stacks, 2)}$ .
- $\texttt{stacks[2]}$ is not empty, set \texttt{new_pos := 2}, and pop the top element of $\texttt{stacks[2]}$ , which makes $\texttt{stacks}$ become $[[1,2],[3,1],[1,2]]$ , and then call $\texttt{get(stacks, 2)}$ .
- $\texttt{stacks[2]}$ is not empty, set \texttt{new_pos := 1}, and pop the top element of $\texttt{stacks[2]}$ , which makes $\texttt{stacks}$ become $[[1,2],[3],[1,2]]$ , and then call $\texttt{get(stacks, 1)}$ .
- $\texttt{stacks[1]}$ is not empty, set \texttt{new_pos := 2}, and pop the top element of $\texttt{stacks[1]}$ , which makes $\texttt{stacks}$ become $[[1],[3],[1,2]]$ , and then call $\texttt{get(stacks, 2)}$ .
- $\texttt{stacks[2]}$ is not empty, set \texttt{new_pos := 3}, and pop the top element of $\texttt{stacks[2]}$ , which makes $\texttt{stacks}$ become $[[1],[],[1,2]]$ , and then call $\texttt{get(stacks, 3)}$ .
- $\texttt{stacks[3]}$ is not empty, set \texttt{new_pos := 2}, and pop the top element of $\texttt{stacks[3]}$ , which makes $\texttt{stacks}$ become $[[1],[],[1]]$ , and then call $\texttt{get(stacks, 2)}$ .
- $\texttt{stacks[2]}$ is empty, return $2$ .

在第一个例子中：

当调用 $\texttt{init(1)}$ $init(1)$ 时，设置 $\texttt{stacks := [[1,2,2],[3,1,2],[1,2,1]]}$ $stacks := [[1,2,2],[3,1,2],[1,2,1]]$ ，然后调用 $\texttt{get(stacks, 1)}$ $get(stacks, 1)$ 。
- $\texttt{stacks[1]}$ 非空，设 \texttt{new_pos := 2}，并弹出 $\texttt{stacks[1]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1,2],[3,1,2],[1,2,1]]$ ，然后调用 $\texttt{get(stacks, 2)}$ 。
- $\texttt{stacks[2]}$ 非空，设 \texttt{new_pos := 2}，并弹出 $\texttt{stacks[2]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1,2],[3,1],[1,2,1]]$ ，然后调用 $\texttt{get(stacks, 2)}$ 。
- $\texttt{stacks[2]}$ 非空，设 \texttt{new_pos := 1}，并弹出 $\texttt{stacks[2]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1,2],[3],[1,2,1]]$ ，然后调用 $\texttt{get(stacks, 1)}$ 。
- $\texttt{stacks[1]}$ 非空，设 \texttt{new_pos := 2}，并弹出 $\texttt{stacks[1]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1],[3],[1,2,1]]$ ，然后调用 $\texttt{get(stacks, 2)}$ 。
- $\texttt{stacks[2]}$ 非空，设 \texttt{new_pos := 3}，并弹出 $\texttt{stacks[2]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1],[],[1,2,1]]$ ，然后调用 $\texttt{get(stacks, 3)}$ 。
- $\texttt{stacks[3]}$ 非空，设 \texttt{new_pos := 1}，并弹出 $\texttt{stacks[3]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1],[],[1,2]]$ ，然后调用 $\texttt{get(stacks, 1)}$ 。
- $\texttt{stacks[1]}$ 非空，设 \texttt{new_pos := 1}，并弹出 $\texttt{stacks[1]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[],[],[1,2]]$ ，然后调用 $\texttt{get(stacks, 1)}$ 。
- $\texttt{stacks[1]}$ 为空，返回 $1$ 。
当调用 $\texttt{init(2)}$ $init(2)$ 时，设置 $\texttt{stacks := [[1,2,2],[3,1,2],[1,2,1]]}$ $stacks := [[1,2,2],[3,1,2],[1,2,1]]$ ，然后调用 $\texttt{get(stacks, 2)}$ $get(stacks, 2)$ 。
- $\texttt{stacks[2]}$ 非空，设 \texttt{new_pos := 2}，并弹出 $\texttt{stacks[2]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1,2,2],[3,1],[1,2,1]]$ ，然后调用 $\texttt{get(stacks, 2)}$ 。
- $\texttt{stacks[2]}$ 非空，设 \texttt{new_pos := 1}，并弹出 $\texttt{stacks[2]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1,2,2],[3],[1,2,1]]$ ，然后调用 $\texttt{get(stacks, 1)}$ 。
- $\texttt{stacks[1]}$ 非空，设 \texttt{new_pos := 2}，并弹出 $\texttt{stacks[1]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1,2],[3],[1,2,1]]$ ，然后调用 $\texttt{get(stacks, 2)}$ 。
- $\texttt{stacks[2]}$ 非空，设 \texttt{new_pos := 3}，并弹出 $\texttt{stacks[2]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1,2],[],[1,2,1]]$ ，然后调用 $\texttt{get(stacks, 3)}$ 。
- $\texttt{stacks[3]}$ 非空，设 \texttt{new_pos := 1}，并弹出 $\texttt{stacks[3]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1,2],[],[1,2]]$ ，然后调用 $\texttt{get(stacks, 1)}$ 。
- $\texttt{stacks[1]}$ 非空，设 \texttt{new_pos := 2}，并弹出 $\texttt{stacks[1]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1],[],[1,2]]$ ，然后调用 $\texttt{get(stacks, 2)}$ 。
- $\texttt{stacks[2]}$ 为空，返回 $2$ 。
当调用 $\texttt{init(3)}$ $init(3)$ 时，设置 $\texttt{stacks := [[1,2,2],[3,1,2],[1,2,1]]}$ $stacks := [[1,2,2],[3,1,2],[1,2,1]]$ ，然后调用 $\texttt{get(stacks, 3)}$ $get(stacks, 3)$ 。
- $\texttt{stacks[3]}$ 非空，设 \texttt{new_pos := 1}，并弹出 $\texttt{stacks[3]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1,2,2],[3,1,2],[1,2]]$ ，然后调用 $\texttt{get(stacks, 1)}$ 。
- $\texttt{stacks[1]}$ 非空，设 \texttt{new_pos := 2}，并弹出 $\texttt{stacks[1]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1,2],[3,1,2],[1,2]]$ ，然后调用 $\texttt{get(stacks, 2)}$ 。
- $\texttt{stacks[2]}$ 非空，设 \texttt{new_pos := 2}，并弹出 $\texttt{stacks[2]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1,2],[3,1],[1,2]]$ ，然后调用 $\texttt{get(stacks, 2)}$ 。
- $\texttt{stacks[2]}$ 非空，设 \texttt{new_pos := 1}，并弹出 $\texttt{stacks[2]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1,2],[3],[1,2]]$ ，然后调用 $\texttt{get(stacks, 1)}$ 。
- $\texttt{stacks[1]}$ 非空，设 \texttt{new_pos := 2}，并弹出 $\texttt{stacks[1]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1],[3],[1,2]]$ ，然后调用 $\texttt{get(stacks, 2)}$ 。
- $\texttt{stacks[2]}$ 非空，设 \texttt{new_pos := 3}，并弹出 $\texttt{stacks[2]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1],[],[1,2]]$ ，然后调用 $\texttt{get(stacks, 3)}$ 。
- $\texttt{stacks[3]}$ 非空，设 \texttt{new_pos := 2}，并弹出 $\texttt{stacks[3]}$ 的栈顶元素，使 $\texttt{stacks}$ 变为 $[[1],[],[1]]$ ，然后调用 $\texttt{get(stacks, 2)}$ 。
- $\texttt{stacks[2]}$ 为空，返回 $2$ 。

CF1889D.Game of Stacks

题目描述

输入格式

输出格式

输入输出样例

说明/提示