codeforces 361 B - Mike and Shortcuts

Time Limit:3000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

 

Description

Recently, Mike was very busy with studying for exams and contests. Now he is going to chill a bit by doing some sight seeing in the city.

City consists of n intersections numbered from 1 to n. Mike starts walking from his house located at the intersection number 1 and goes along some sequence of intersections. Walking from intersection number i to intersection j requires |i - j| units of energy. The total energyspent by Mike to visit a sequence of intersections p₁ = 1, p₂, ..., p_k is equal to  units of energy.

Of course, walking would be boring if there were no shortcuts. A shortcut is a special path that allows Mike walking from one intersection to another requiring only 1 unit of energy. There are exactly n shortcuts in Mike's city, the i^th of them allows walking from intersection i to intersection a_i (i ≤ a_i ≤ a_i + 1) (but not in the opposite direction), thus there is exactly one shortcut starting at each intersection. Formally, if Mike chooses a sequence p₁ = 1, p₂, ..., p_k then for each 1 ≤ i < k satisfying p_i + 1 = a_pi and a_pi ≠ p_i Mike will spend only 1 unit of energy instead of |p_i - p_i + 1| walking from the intersection p_i to intersection p_i + 1. For example, if Mike chooses a sequence p₁ = 1, p₂ = a_p1, p₃ = a_p2, ..., p_k = a_pk - 1, he spends exactly k - 1 units of total energy walking around them.

Before going on his adventure, Mike asks you to find the minimum amount of energy required to reach each of the intersections from his home. Formally, for each 1 ≤ i ≤ n Mike is interested in finding minimum possible total energy of some sequence p₁ = 1, p₂, ..., p_k = i.

Input

The first line contains an integer n(1 ≤ n ≤ 200 000) — the number of Mike's city intersection.

The second line contains n integers a₁, a₂, ..., a_n(i ≤ a_i ≤ n , , describing shortcuts of Mike's city, allowing to walk from intersection i to intersection a_i using only 1 unit of energy. Please note that the shortcuts don't allow walking in opposite directions (from a_i to i).

Output

In the only line print n integers m₁, m₂, ..., m_n, where m_i denotes the least amount of total energy required to walk from intersection 1 to intersection i.

Sample Input

Input

3 2 2 3

Output

0 1 2 

Input

5 1 2 3 4 5

Output

0 1 2 3 4 

Input

7 4 4 4 4 7 7 7

Output

0 1 2 1 2 3 3 

Hint

In the first sample case desired sequences are:

1: 1; m₁ = 0;

2: 1, 2; m₂ = 1;

3: 1, 3; m₃ = |3 - 1| = 2.

In the second sample case the sequence for any intersection 1 < i is always 1, i and m_i = |1 - i|.

In the third sample case — consider the following intersection sequences:

1: 1; m₁ = 0;

2: 1, 2; m₂ = |2 - 1| = 1;

3: 1, 4, 3; m₃ = 1 + |4 - 3| = 2;

4: 1, 4; m₄ = 1;

5: 1, 4, 5; m₅ = 1 + |4 - 5| = 2;

6: 1, 4, 6; m₆ = 1 + |4 - 6| = 3;

7: 1, 4, 5, 7; m₇ = 1 + |4 - 5| + 1 = 3.

 

n个城市排成一排，起点是第一个城市，只有三种情况，i，i+1,i-1,直接用bfs求最短路径

 

#include<bits/stdc++.h>
using namespace std;
int num[200005], ans[2000005];
int main()
{
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
        scanf("%d", num+i);
    memset(ans, -1, sizeof(ans));
    ans[1] = 0;
    queue<int> q;
    q.push(1);
    while(!q.empty())
    {
        int s = q.front();
        q.pop();
        for(int i = -1; i <= 1; ++i)
        {
            if(i == 0)
                continue;
            int d = s + i;
            if(d >= 1 && d <= n && ans[d] == -1)
            {
                ans[d] = ans[s] + 1;
                q.push(d);
            }
        }
        int d = num[s];
        if(ans[d] == -1)
        {
            ans[d] = ans[s] + 1;
            q.push(d);
        }
    }
    printf("%d", ans[1]);
    for(int i = 2; i <= n; i++)
        printf(" %d", ans[i]);
    cout<<endl;
    return 0;
}