CodeForces 689B Mike and Shortcuts （bfs or 最短路）

Mike and Shortcuts

题目链接：

http://acm.hust.edu.cn/vjudge/contest/121333#problem/F

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 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 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 sequence p1 = 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.

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 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.

题意：

图中任意两点之间距离为abs(i-j)，每个点有且仅有一条捷径a[i]并且距离为1；
求从1开始遍历所有点的最小代价；

题解：

只考虑相邻两点和捷径这三条长度为1的边，最短路即可；
或者直接考虑上述三条边跑一遍bfs.
（以下代码为bfs法）

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<set>
#include<list>
#define LL long long
#define maxn 210000
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
using namespace std;

int n;
queue<int> myq;
int dis[maxn];
int a[maxn];

int main(int argc, char const *argv[])
{
   //IN;

    while(scanf("%d",&n) != EOF)
    {
        for(int i=1; i<=n; i++) scanf("%d", &a[i]);
        fill(dis, dis+maxn, inf);
        while(!myq.empty()) myq.pop();

        myq.push(1); dis[1] = 0;
        while(!myq.empty()) {
            int cur = myq.front(); myq.pop();

            if(cur+1<=n && dis[cur+1] > dis[cur]+1){
                dis[cur+1] = dis[cur]+1;
                myq.push(cur+1);
            }
            if(cur-1>0 && dis[cur-1] > dis[cur]+1){
                dis[cur-1] = dis[cur]+1;
                myq.push(cur-1);
            }
            if(dis[a[cur]] > dis[cur]+1){
                dis[a[cur]] = dis[cur]+1;
                myq.push(a[cur]);
            }
        }

        for(int i=1; i<=n; i++)
            printf("%d ", dis[i]);;
    }

    return 0;
}