codeforces 689B B. Mike and Shortcuts(bfs)

题目链接：

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, ..., pk 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 ith of them allows walking from intersection ito intersection ai (i ≤ ai ≤ ai + 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, ..., pk then for each 1 ≤ i < k satisfying pi + 1 = api and api ≠ pi Mike will spend only 1unit of energy instead of |pi - pi + 1| walking from the intersection pi to intersection pi + 1. For example, if Mike chooses a sequencep1 = 1, p2 = ap1, p3 = ap2, ..., pk = apk - 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, ..., pk = 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, ..., an (i ≤ ai ≤ n , , describing shortcuts of Mike's city, allowing to walk from intersection i to intersection ai using only 1 unit of energy. Please note that the shortcuts don't allow walking in opposite directions (from ai to i).

Output

 

In the only line print n integers m1, m2, ..., mn, where mi 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 

题意：

点i与点j的距离为|i-j|,而且第i个点到a[i]的距离为1,问第1个点到其他所有点的最短距离;

思路：

bfs的水题,对于一个点i,距它为1的点有i-1,i+1，a[i],这样bfs就好了;

AC代码：

//#include <bits/stdc++.h>
#include <vector>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <cstring>
#include <algorithm>
#include <cstdio>

using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef  long long LL;
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=2e5+10;
const int maxn=1005;
const double eps=1e-10;

int n,a[N],vis[N],dis[N];
queue<int>qu;
void bfs()
{
    qu.push(1);
    vis[1]=1;
    while(!qu.empty())
    {
        int fr=qu.front();
        qu.pop();
        if(!vis[fr+1])
        {
            vis[fr+1]=1;
            dis[fr+1]=dis[fr]+1;
            qu.push(fr+1);
        }
        if(fr-1>0&&!vis[fr-1])
        {  vis[fr-1]=1;
            dis[fr-1]=dis[fr]+1;
            qu.push(fr-1);

        }
        if(!vis[a[fr]])
        {
            vis[a[fr]]=1;
            dis[a[fr]]=dis[fr]+1;
            qu.push(a[fr]);
        }
    }
}
int main()
{
    read(n);
    For(i,1,n)read(a[i]);
    bfs();
    For(i,1,n)printf("%d ",dis[i]);

        return 0;
}