Codeforces Round #361 (Div. 2) B bfs处理

B. Mike and Shortcuts

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 energy spent 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.

Examples

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 

Note

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.

题意：  从位置1出发到其他位置，距离为|i - j|    但是存在近路一说，也就是可以直接从i到ai 距离为1

输出从位置1到其他位置的最少距离

题解：最少距离bfs处理

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<queue>
 using namespace std;
 struct node
 {
  int pos;
  int step;
 }N[],now,exm;
 int n;
 int a[];
 int b[];
 int ans[];
 queue<node>q;
 int main()
 {
     scanf("%d",&n);
     memset(b,,sizeof(b));
     for(int i=;i<=n;i++)
     N[i].pos=i;
     for(int i=;i<=n;i++)
       scanf("%d",&a[i]);
     exm.pos=;
     exm.step=;
     b[]=;
     q.push(exm);

     while(!q.empty())
     {
         exm=q.front();
        // cout<<exm.pos<<" "<<exm.step<<endl;
         ans[exm.pos]=exm.step;
         q.pop();
         for(int i=;i<=;i++)
         {
             if(i==)
             {
                if(b[exm.pos+]==&&exm.pos+<=n)
                {
                    now.pos=exm.pos+;
                    now.step=exm.step+;
                    q.push(now);
                    b[now.pos]=;
                }
             }
             if(i==)
             {
                if(b[exm.pos-]==&&exm.pos->=)
                {
                    now.pos=exm.pos-;
                    now.step=exm.step+;
                    q.push(now);
                    b[now.pos]=;
                }
             }
             if(i==)
             {
                 if(b[a[exm.pos]]==&&a[exm.pos]>=&&a[exm.pos]<=n)
                 {
                     now.pos=a[exm.pos];
                     now.step=exm.step+;
                     q.push(now);
                     b[a[exm.pos]]=;
                 }
             }
         }
     }
     cout<<ans[];
     for(int i=;i<=n;i++)
         printf(" %d",ans[i]);
     return ;
 }