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A16353. Game of Stacks

编程题 普及/提高-
知识点

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 $\texttt{init(1)}$ , set $\texttt{stacks := [[1,2,2],[3,1,2],[1,2,1]]}$ , 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 $[[1,2],[3,1,2],[1,2,1]]$ , 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 $\texttt{init(2)}$ , set $\texttt{stacks := [[1,2,2],[3,1,2],[1,2,1]]}$ , 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 $[[1,2,2],[3,1],[1,2,1]]$ , 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 $\texttt{init(3)}$ , set $\texttt{stacks := [[1,2,2],[3,1,2],[1,2,1]]}$ , 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 $[[1,2,2],[3,1,2],[1,2]]$ , 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 $2$ .
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