CSU-ACM2017暑期训练6-bfs G

题目：

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 intersectionp_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.

Example

Input

32 2 3

Output

0 1 2 

Input

51 2 3 4 5

Output

0 1 2 3 4 

Input

74 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, iand 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.

              题目大意：给你n个点,每个点到另一个点的消耗为两点坐标的差的绝对值。每个点还有一条到另一个点的绝对值，从这个点到那个点消耗为1，捷径是单向的，问你从1到n每个点到1这个点的最少消耗

             思路:相当于对一个图的每个点求他们到点1的单源最短距离。这个图中的路就是每个点比如点i到i-1和i+1有一条长为1的路，还有一条到a[i]的路长也为1.于是这题就变成了直接bfs搜从点1到各个点的最短距离

代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<list>#include<deque>#include<map>#include<queue>using namespace std;typedef long long ll;const double PI = acos(-1.0);const double eps = 1e-6;const int INF = 1000000000;const int maxn = 222222;int T,n,m;int a[222222];int d[222222];int vis[222222];queue<int>q;void bfs(int x){    q.push(x);    while(!q.empty())    {        int t=q.front();        q.pop();        if(vis[t]==1)            continue;        vis[t]=1;        for(int i=-1;i<=1;i++)        {            if(i==0)            {                if(d[t]+1<d[a[t]])                    d[a[t]]=d[t]+1;                q.push(a[t]);            }            else            {                if(t+i>0&&t+i<=n&&d[t]+1<d[t+i])                {                    d[t+i]=d[t]+1;                    q.push(t+i);                }            }        }    }}int main(){    scanf("%d",&n);    for(int i=0;i<=n;i++)        d[i]=INF;    d[1] = 0;    for(int i=1;i<=n;i++)    {        scanf("%d",&a[i]);    }    bfs(1);    printf("%d",d[1]);    for(int i=2;i<=n;i++)        printf(" %d",d[i]);    return 0;}