F. Selling a Menagerie

F. Selling a Menagerie

You are the owner of a menagerie consisting of $n$ animals numbered from $1$ to $n$. However, maintaining the menagerie is quite expensive, so you have decided to sell it!

It is known that each animal is afraid of exactly one other animal. More precisely, animal $i$ is afraid of animal $a_i$ ($a_i \neq i$). Also, the cost of each animal is known, for animal $i$ it is equal to $c_i$.

You will sell all your animals in some fixed order. Formally, you will need to choose some permutation$^\dagger$ $p_1, p_2, \ldots, p_n$, and sell animal $p_1$ first, then animal $p_2$, and so on, selling animal $p_n$ last.

When you sell animal $i$, there are two possible outcomes:

• If animal $a_i$ was sold before animal $i$, you receive $c_i$ money for selling animal $i$.
• If animal $a_i$ was not sold before animal $i$, you receive $2 \cdot c_i$ money for selling animal $i$. (Surprisingly, animals that are currently afraid are more valuable).

Your task is to choose the order of selling the animals in order to maximize the total profit.

For example, if $a = [3, 4, 4, 1, 3]$, $c = [3, 4, 5, 6, 7]$, and the permutation you choose is $[4, 2, 5, 1, 3]$, then:

• The first animal to be sold is animal $4$. Animal $a_4 = 1$ was not sold before, so you receive $2 \cdot c_4 = 12$ money for selling it.
• The second animal to be sold is animal $2$. Animal $a_2 = 4$ was sold before, so you receive $c_2 = 4$ money for selling it.
• The third animal to be sold is animal $5$. Animal $a_5 = 3$ was not sold before, so you receive $2 \cdot c_5 = 14$ money for selling it.
• The fourth animal to be sold is animal $1$. Animal $a_1 = 3$ was not sold before, so you receive $2 \cdot c_1 = 6$ money for selling it.
• The fifth animal to be sold is animal $3$. Animal $a_3 = 4$ was sold before, so you receive $c_3 = 5$ money for selling it.

Your total profit, with this choice of permutation, is $12 + 4 + 14 + 6 + 5 = 41$. Note that $41$ is not the maximum possible profit in this example.

$^\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in any order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array) and $[1,3,4]$ is also not a permutation ($n=3$, but $4$ is present in the array).

Input

The first line of the input contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

Then follow the descriptions of the test cases.

The first line of each test case description contains an integer $n$ ($2 \le n \le 10^5$) — the number of animals.

The second line of the test case description contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le n$, $a_i \neq i$) — $a_i$ means the index of the animal that animal $i$ is afraid of.

The third line of the test case description contains $n$ integers $c_1, c_2, \dots, c_n$ ($1 \le c_i \le 10^9$) — the costs of the animals.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output

Output $t$ lines, each containing the answer to the corresponding test case. The answer should be $n$ integers — the permutation $p_1, p_2, \ldots, p_n$, indicating in which order to sell the animals in order to maximize the profit. If there are multiple possible answers, you can output any of them.

Example

input

8
3
2 3 2
6 6 1
8
2 1 4 3 6 5 8 7
1 2 1 2 2 1 2 1
5
2 1 1 1 1
9 8 1 1 1
2
2 1
1000000000 999999999
7
2 3 2 6 4 4 3
1 2 3 4 5 6 7
5
3 4 4 1 3
3 4 5 6 7
3
2 1 1
1 2 2
4
2 1 4 1
1 1 1 1

output

1 2 3
2 4 5 1 6 3 7 8
3 4 5 1 2
1 2
7 5 1 3 2 6 4
5 3 2 4 1
3 2 1
3 4 1 2

 

解题思路

　　比赛最后几分钟做了出来，主要卡在了处理挂在环上的链，代码写得又长又臭。

　　如果从$i$到$a_i$连一条边$i \to a_i$，那么可以发现得到的图是一个基环树，这是因为每个节点都恰好有一个出度。分别考虑基环树中的环以及挂在环上的树。对于挂在环上的树，也就是不在环中的点，我们总是可以通过拓扑排序找出来，那么剩余的点自然就是环上的点了，这样就可以把环以及挂在环上的树分离开来。

　　对于某个节点$u$，如果存在害怕$u$的动物$v$，即$u$的入度不为$0$，那么很明显先卖$v$再卖$u$所得到的收益比先卖$u$再卖$v$要多（至少不会变差）。因此继续顺着找到害怕$v$的动物，以此类推直到找到某个点$x$，不存在害怕$x$的动物，即$x$的入度为$0$，那么就先售卖$x$，这样做肯定是最优的。

　　为此先对不在环中的点进行售卖，可以发现每个点都能获得两倍的收益。只需按照拓扑的顺序进行售卖即可，这样就可以保证每次卖掉的点的入度均为$0$。然后就是环上的点，可以发现无论按什么顺序卖掉，都至少有一个点不能获得两倍收益（因为必然存在只剩一个点的情况）。为了能得到最大收益，假设环中$c_x$最小，那么我们按照$a_x \to a_{a_x} \to \cdots \to x$的顺序进行售卖，那么除了$x$外，其余的点都能获得两倍收益。

　　要注意得到的图不一定是连通的，因此要分别处理每个基环树。

　　AC代码如下，时间复杂度为$O(n)$：

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N = 1e5 + 10;

int a[N], c[N];
int deg[N];
bool vis[N];
int q[N], hh, tt;

void solve() {
    int n;
    scanf("%d", &n);
    memset(deg, 0, n + 10 << 2);
    memset(vis, 0, n + 10);
    for (int i = 1; i <= n; i++) {
        scanf("%d", a + i);
        deg[a[i]]++;
    }
    for (int i = 1; i <= n; i++) {
        scanf("%d", c + i);
    }
    hh = 0, tt = -1;
    for (int i = 1; i <= n; i++) {
        if (!deg[i]) q[++tt] = i;
    }
    while (hh <= tt) {    // 先把不在环上的点找出来 
        int t = q[hh++];
        vis[t] = true;    // 不在环上的点被标记 
        printf("%d ", t);    // t的入度为0，因此卖掉 
        if (--deg[a[t]] == 0) q[++tt] = a[t];
    }
    for (int i = 1; i <= n; i++) {
        if (vis[i]) continue;    // 没有被标记的点一定是环上的点 
        int t = i;
        vector<int> p;
        while (!vis[t]) {    // 把环上的点都找出来 
            vis[t] = true;
            p.push_back(t);
            t = a[t];
        }
        int k = 0;    // 找到收益最小的点 
        for (int i = 0; i < p.size(); i++) {
            if (c[p[i]] < c[p[k]]) k = i;
        }
        for (int i = 0; i < p.size(); i++) {    // 收益最小的点最后卖掉 
            if (++k == p.size()) k = 0;
            printf("%d ", p[k]);
        }
    }
    printf("\n");
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        solve();
    }
    
    return 0;
}

 

参考资料

　　Codeforces Round 895 (Div. 3) Editorial：https://codeforces.com/blog/entry/120165