CF1889D.Game of Stacks

普及/提高-

通过率：0%

AC君温馨提醒

该题目为【codeforces】题库的题目，您提交的代码将被提交至codeforces进行远程评测，并由ACGO抓取测评结果后进行展示。由于远程测评的测评机由其他平台提供，我们无法保证该服务的稳定性，若提交后无反应，请等待一段时间后再进行重试。

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.

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, $\sum k_i\le 10^6$ .

输出格式

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

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 $2$ .

首页