Codeforces Round #361 (Div. 2)——B. Mike and Shortcuts（BFS+小坑）

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 energyspent 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 i to 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 1 unit 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 

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 mi = |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.

这题基本不能用SPFA，20W个点，边多到存不下，SPFA会WA4，然后试了下BFS，也WA4，发现有个坑，第四组中到第10个点的最短距离是8，是1走捷径到17，再一步一步往回走到10即1+7=8而不是直接从1走到10的直接距离9，显然这里需要往回走。BFS里面多写一个往回走的方向搜索就好了

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define MM(x,y) memset(x,y,sizeof(x))
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=200010;

int vis[N];
int cut[N];
int d[N];
queue<pii>Q;

void init()
{
	MM(vis,0);
	MM(cut,-1);
	MM(d,0);
	while (!Q.empty())
		Q.pop();
}
void bfs(const int &s)
{
	Q.push(pii(s,0));
	while (!Q.empty())
	{
		pii now=Q.front();
		Q.pop();
		vis[now.first]=1;
		d[now.first]=now.second;
		int v;
		v=cut[now.first];
		if(!vis[v]&&v!=-1)
		{
			Q.push(pii(v,now.second+1));
			vis[v]=1;
		}	
		v=now.first+1;		
		if(!vis[v])
		{
			Q.push(pii(v,now.second+1));
			vis[v]=1;
		}
		v=now.first-1;
		if(!vis[v])
		{
			Q.push(pii(v,now.second+1));
			vis[now.first-1]=1;
		}	
	}
	return ;
}
int main(void)
{
	int n,i,j,a,b;
	while (~scanf("%d",&n))
	{
		init();
		for (i=1; i<=n; i++)
		{
			scanf("%d",&a);
			cut[i]=a;
		}
		bfs(1);
		for (i=1; i<=n; i++)
			printf("%d%s",d[i],i==n?"
":" ");
	}
	return 0;
}