Codeforces 34D. Road Map 树的遍历

D. Road Map

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n cities in Berland. Each city has its index — an integer number from 1 to n. The capital has index r₁. All the roads in Berland are two-way. The road system is such that there is exactly one path from the capital to each city, i.e. the road map looks like a tree. In Berland's chronicles the road map is kept in the following way: for each city i, different from the capital, there is kept number p_i — index of the last city on the way from the capital to i.

Once the king of Berland Berl XXXIV decided to move the capital from city r₁ to city r₂. Naturally, after this the old representation of the road map in Berland's chronicles became incorrect. Please, help the king find out a new representation of the road map in the way described above.

Input

The first line contains three space-separated integers n, r₁, r₂ (2 ≤ n ≤ 5·10⁴, 1 ≤ r₁ ≠ r₂ ≤ n) — amount of cities in Berland, index of the old capital and index of the new one, correspondingly.

The following line contains n - 1 space-separated integers — the old representation of the road map. For each city, apart from r₁, there is given integer p_i — index of the last city on the way from the capital to city i. All the cities are described in order of increasing indexes.

Output

Output n - 1 numbers — new representation of the road map in the same format.

Sample test(s)

input

3 2 3
2 2

output

2 3 

input

6 2 4
6 1 2 4 2

output

6 4 1 4 2 

给一棵树，换个顶点遍历一次。

#include <iostream>
#include <cstring>
#include <vector>

using namespace std;

int n,r1,r2;
vector<int>a[111111];
int p[111111]={0};
bool v[111111]={0};

void dfs(int i)
{
    v[i]=true;
    for (int j=0;j<a[i].size();j++)
    {
        if (!v[a[i][j]])
        {
            p[a[i][j]]=i;
            dfs(a[i][j]);
        }
    }
}

int main()
{
    cin>>n>>r1>>r2;
    for (int i=1;i<=n;i++)
    {
        if (i!=r1)
        {
            int j;
            cin>>j;
            a[i].push_back(j);
            a[j].push_back(i);
        }
    }
    dfs(r2);
    for (int i=1;i<=n;i++)
    {
        if (i!=r2)
        {
            cout<<p[i]<<" ";
        }
    }
    cout<<endl;
    return 0;
}