Buggy python submission (verdict: Time Limit Exceeded):
import heapq
import collections
 
N,X,Y = (int(i) for i in input().split())
 
graph = collections.defaultdict(list)
# X->Yの経路を追加
graph[X].append((1,Y))
graph[Y].append((1,X))
# grpah[node_num] = [(cost,next_node_num)]
for i in range(1,N):
  graph[i].append((1,i+1))
  graph[i+1].append((1,i))
 
# ans[i] = 最短経路がiである個数
ans = [0] * (N-1)
# Node 1 ~ N Nodeまでをループ
for i in range(1,N+1):
  # priority_queue : (cost, node_num)
  priority_queue = [(0,i)]
  # 探索したnode
  checked_node = {}
 
  while priority_queue:
    (cost, node_num) = heapq.heappop(priority_queue)
    if node_num in checked_node:
      continue
    else:
      if cost != 0:
        ans[cost - 1] += 1
      checked_node[node_num] = cost
      for next_cost,next_node_num in graph[node_num]:
        if next_node_num not in checked_node:
          heapq.heappush(priority_queue, (cost + next_cost, next_node_num))

for a in ans:
  print(a//2)

Fixed submission:
# -*- coding: utf-8 -*-
from collections import Counter

def update_distance(d, num, N):
    minn = num
    maxx = num
    # 要素数の小さい方に向けて
    while True:
        if minn > 0:
            minn -= 1
        else:
            break
        if d[minn] > d[minn+1] + 1:
            d[minn] = d[minn+1] + 1
        else:
            break
    # 要素数の大きいほうに向けて
    while True:
        if maxx < (N - 1):
            maxx += 1
        else:
            break
        if d[maxx] > d[maxx-1] + 1:
            d[maxx] = d[maxx-1] + 1
        else:
            break

(N, X, Y) = (int(i) for i in input().split())
X = X-1
Y = Y-1

# まずは経路情報をd[N][N]の二次元配列に格納
# (i, j)がつながっていれば1, つながってなければ2*10^3

d = []
for i in range(N):
    dd = []
    for j in range(N):
        dd.append(abs(i - j))
    d.append(dd)

# XとYがつながることで最短経路を修正していく
for i in range(N):
    dd = d[i]
    dd[X] = min(dd[X], dd[Y] + 1)
    dd[Y] = min(dd[Y], dd[X] + 1)
    update_distance(dd, X, N)
    update_distance(dd, Y, N)

# dの全リストを連結してあとは0~N-1までカウントする
d_list = []
for i in range(N - 1):
    d_list += d[i][i+1:]

#print(d_list)
c = Counter(d_list)

for i in range(1, N):
    if i in c:
        print(c[i])
    else:
        print(0)

Buggy python submission (verdict: Time Limit Exceeded):
import heapq
import collections
 
N,X,Y = (int(i) for i in input().split())
 
graph = collections.defaultdict(list)
# X->Yの経路を追加
graph[X].append((1,Y))
graph[Y].append((1,X))
# grpah[node_num] = [(cost,next_node_num)]
for i in range(1,N):
  graph[i].append((1,i+1))
  graph[i+1].append((1,i))
 
# ans[i] = 最短経路がiである個数
ans = [0] * (N-1)
# Node 1 ~ N Nodeまでをループ
for i in range(1,N+1):
  # priority_queue : (cost, node_num)
  priority_queue = [(0,i)]
  # 探索したnode
  checked_node = {}
 
  while priority_queue:
    (cost, node_num) = heapq.heappop(priority_queue)
    if node_num in checked_node:
      continue
    else:
      checked_node[node_num] = cost
      for next_cost,next_node_num in graph[node_num]:
        if next_node_num not in checked_node:
          heapq.heappush(priority_queue, (cost + next_cost, next_node_num))
  values = checked_node.values()
  c = collections.Counter(values)
  for k, v in c.items():
      if k != 0:
          ans[k - 1] += v
 
for a in ans:
  print(a//2)

Fixed submission:
# -*- coding: utf-8 -*-
from collections import Counter

def update_distance(d, num, N):
    minn = num
    maxx = num
    # 要素数の小さい方に向けて
    while True:
        if minn > 0:
            minn -= 1
        else:
            break
        if d[minn] > d[minn+1] + 1:
            d[minn] = d[minn+1] + 1
        else:
            break
    # 要素数の大きいほうに向けて
    while True:
        if maxx < (N - 1):
            maxx += 1
        else:
            break
        if d[maxx] > d[maxx-1] + 1:
            d[maxx] = d[maxx-1] + 1
        else:
            break

(N, X, Y) = (int(i) for i in input().split())
X = X-1
Y = Y-1

# まずは経路情報をd[N][N]の二次元配列に格納
# (i, j)がつながっていれば1, つながってなければ2*10^3

d = []
for i in range(N):
    dd = []
    for j in range(N):
        dd.append(abs(i - j))
    d.append(dd)

# XとYがつながることで最短経路を修正していく
for i in range(N):
    dd = d[i]
    dd[X] = min(dd[X], dd[Y] + 1)
    dd[Y] = min(dd[Y], dd[X] + 1)
    update_distance(dd, X, N)
    update_distance(dd, Y, N)

# dの全リストを連結してあとは0~N-1までカウントする
d_list = []
for i in range(N - 1):
    d_list += d[i][i+1:]

#print(d_list)
c = Counter(d_list)

for i in range(1, N):
    if i in c:
        print(c[i])
    else:
        print(0)

Buggy python submission (verdict: Time Limit Exceeded):
# -*- coding: utf-8 -*-
from collections import Counter
 
(N, X, Y) = (int(i) for i in input().split())
 
# まずは経路情報をd[N][N]の二次元配列に格納
# (i, j)がつながっていれば1, つながってなければ2*10^3
 
d = []
for i in range(N):
    d.append([2 * 10**3]*N)
 
for i in range(N - 1):
    d[i][i] = 0
    d[i][i+1] = 1
    d[i+1][i] = 1
 
d[X-1][Y-1] = 1
d[Y-1][X-1] = 1
#print(d)
 
# ワーシャル・フロイドで全点の最短経路を計算
for k in range(N):
    for i in range(N):
        for j in range(N):
            d[i][j] = min(d[i][j], d[i][k] + d[k][j])
 
#print(d)

# dの全リストを連結してあとは0~N-1までカウントする
d_list = []
for i in range(N - 1):
    d_list += d[i][i+1:]

#print(d_list)
c = Counter(d_list)

for i in range(1, N):
    if i in c:
        print(c[i])
    else:
        print(0)

Fixed submission:
# -*- coding: utf-8 -*-
from collections import Counter

def update_distance(d, num, N):
    minn = num
    maxx = num
    # 要素数の小さい方に向けて
    while True:
        if minn > 0:
            minn -= 1
        else:
            break
        if d[minn] > d[minn+1] + 1:
            d[minn] = d[minn+1] + 1
        else:
            break
    # 要素数の大きいほうに向けて
    while True:
        if maxx < (N - 1):
            maxx += 1
        else:
            break
        if d[maxx] > d[maxx-1] + 1:
            d[maxx] = d[maxx-1] + 1
        else:
            break

(N, X, Y) = (int(i) for i in input().split())
X = X-1
Y = Y-1

# まずは経路情報をd[N][N]の二次元配列に格納
# (i, j)がつながっていれば1, つながってなければ2*10^3

d = []
for i in range(N):
    dd = []
    for j in range(N):
        dd.append(abs(i - j))
    d.append(dd)

# XとYがつながることで最短経路を修正していく
for i in range(N):
    dd = d[i]
    dd[X] = min(dd[X], dd[Y] + 1)
    dd[Y] = min(dd[Y], dd[X] + 1)
    update_distance(dd, X, N)
    update_distance(dd, Y, N)

# dの全リストを連結してあとは0~N-1までカウントする
d_list = []
for i in range(N - 1):
    d_list += d[i][i+1:]

#print(d_list)
c = Counter(d_list)

for i in range(1, N):
    if i in c:
        print(c[i])
    else:
        print(0)

Buggy python submission (verdict: Time Limit Exceeded):
# -*- coding: utf-8 -*-

def update_distance(d, num, N):
    minn = num
    maxx = num
    # 要素数の小さい方に向けて
    while True:
        if minn > 0:
            minn -= 1
        else:
            break
        if d[minn] > d[minn+1] + 1:
            d[minn] = d[minn+1] + 1
        else:
            break
    # 要素数の大きいほうに向けて
    while True:
        if maxx < (N - 1):
            maxx += 1
        else:
            break
        if d[maxx] > d[maxx-1] + 1:
            d[maxx] = d[maxx-1] + 1
        else:
            break

(N, X, Y) = (int(i) for i in input().split())
X = X-1
Y = Y-1

# まずは経路情報をd[N][N]の二次元配列に格納
# (i, j)がつながっていれば1, つながってなければ2*10^3

d = []
for i in range(N):
    dd = []
    for j in range(N):
        dd.append(abs(i - j))
    d.append(dd)

# XとYがつながることで最短経路を修正していく
for i in range(N):
    dd = d[i]
    dd[X] = min(dd[X], dd[Y] + 1)
    dd[Y] = min(dd[Y], dd[X] + 1)
    update_distance(dd, X, N)
    update_distance(dd, Y, N)

# dの全リストを連結してあとは0~N-1までカウントする
d_list = []
for i in range(N - 1):
    d_list += d[i][i+1:]

#print(d_list)

for i in range(1, N):
    count = d_list.count(i)
    print(count)

Fixed submission:
# -*- coding: utf-8 -*-
from collections import Counter

def update_distance(d, num, N):
    minn = num
    maxx = num
    # 要素数の小さい方に向けて
    while True:
        if minn > 0:
            minn -= 1
        else:
            break
        if d[minn] > d[minn+1] + 1:
            d[minn] = d[minn+1] + 1
        else:
            break
    # 要素数の大きいほうに向けて
    while True:
        if maxx < (N - 1):
            maxx += 1
        else:
            break
        if d[maxx] > d[maxx-1] + 1:
            d[maxx] = d[maxx-1] + 1
        else:
            break

(N, X, Y) = (int(i) for i in input().split())
X = X-1
Y = Y-1

# まずは経路情報をd[N][N]の二次元配列に格納
# (i, j)がつながっていれば1, つながってなければ2*10^3

d = []
for i in range(N):
    dd = []
    for j in range(N):
        dd.append(abs(i - j))
    d.append(dd)

# XとYがつながることで最短経路を修正していく
for i in range(N):
    dd = d[i]
    dd[X] = min(dd[X], dd[Y] + 1)
    dd[Y] = min(dd[Y], dd[X] + 1)
    update_distance(dd, X, N)
    update_distance(dd, Y, N)

# dの全リストを連結してあとは0~N-1までカウントする
d_list = []
for i in range(N - 1):
    d_list += d[i][i+1:]

#print(d_list)
c = Counter(d_list)

for i in range(1, N):
    if i in c:
        print(c[i])
    else:
        print(0)

Buggy python submission (verdict: Time Limit Exceeded):
# -*- coding: utf-8 -*-

(N, X, Y) = (int(i) for i in input().split())

# まずは経路情報をd[N][N]の二次元配列に格納
# (i, j)がつながっていれば1, つながってなければ2*10^3

d = []
for i in range(N):
    d.append([2 * 10**3]*N)

for i in range(N - 1):
    d[i][i] = 0
    d[i][i+1] = 1
    d[i+1][i] = 1

d[X-1][Y-1] = 1
d[Y-1][X-1] = 1
#print(d)

# ワーシャル・フロイドで全点の最短経路を計算
for k in range(N):
    for i in range(N):
        for j in range(N):
            d[i][j] = min(d[i][j], d[i][k] + d[k][j])

#print(d)

# dの全リストを連結してあとは0~N-1までカウントする
d_list = []
for i in range(N - 1):
    d_list += d[i][i+1:]

#print(d_list)

for i in range(1, N):
    count = d_list.count(i)
    print(count)

Fixed submission:
# -*- coding: utf-8 -*-
from collections import Counter

def update_distance(d, num, N):
    minn = num
    maxx = num
    # 要素数の小さい方に向けて
    while True:
        if minn > 0:
            minn -= 1
        else:
            break
        if d[minn] > d[minn+1] + 1:
            d[minn] = d[minn+1] + 1
        else:
            break
    # 要素数の大きいほうに向けて
    while True:
        if maxx < (N - 1):
            maxx += 1
        else:
            break
        if d[maxx] > d[maxx-1] + 1:
            d[maxx] = d[maxx-1] + 1
        else:
            break

(N, X, Y) = (int(i) for i in input().split())
X = X-1
Y = Y-1

# まずは経路情報をd[N][N]の二次元配列に格納
# (i, j)がつながっていれば1, つながってなければ2*10^3

d = []
for i in range(N):
    dd = []
    for j in range(N):
        dd.append(abs(i - j))
    d.append(dd)

# XとYがつながることで最短経路を修正していく
for i in range(N):
    dd = d[i]
    dd[X] = min(dd[X], dd[Y] + 1)
    dd[Y] = min(dd[Y], dd[X] + 1)
    update_distance(dd, X, N)
    update_distance(dd, Y, N)

# dの全リストを連結してあとは0~N-1までカウントする
d_list = []
for i in range(N - 1):
    d_list += d[i][i+1:]

#print(d_list)
c = Counter(d_list)

for i in range(1, N):
    if i in c:
        print(c[i])
    else:
        print(0)