結果
| 問題 |
No.8063 幅優先探索
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2020-05-10 17:14:35 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
RE
(最新)
AC
(最初)
|
| 実行時間 | - |
| コード長 | 4,876 bytes |
| コンパイル時間 | 222 ms |
| コンパイル使用メモリ | 81,920 KB |
| 実行使用メモリ | 90,112 KB |
| 最終ジャッジ日時 | 2024-07-07 22:33:25 |
| 合計ジャッジ時間 | 2,337 ms |
|
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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| ファイルパターン | 結果 |
|---|---|
| sample | RE * 2 |
| other | RE * 9 |
ソースコード
import sys
import heapq
import re
from itertools import permutations
from bisect import bisect_left, bisect_right
from collections import Counter, deque
from fractions import gcd
from math import factorial, sqrt, ceil
from functools import lru_cache, reduce
INF = 1 << 60
MOD = 1000000007
sys.setrecursionlimit(10 ** 7)
# UnionFind
class UnionFind():
def __init__(self, n):
self.n = n
self.parents = [-1] * n
def find(self, x):
if self.parents[x] < 0:
return x
else:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if x == y:
return
if self.parents[x] > self.parents[y]:
x, y = y, x
self.parents[x] += self.parents[y]
self.parents[y] = x
def size(self, x):
return -self.parents[self.find(x)]
def same(self, x, y):
return self.find(x) == self.find(y)
def members(self, x):
root = self.find(x)
return [i for i in range(self.n) if self.find(i) == root]
def roots(self):
return [i for i, x in enumerate(self.parents) if x < 0]
def group_count(self):
return len(self.roots())
def all_group_members(self):
return {r: self.members(r) for r in self.roots()}
def __str__(self):
return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())
def is_prime(n):
if n == 1:
return False
for i in range(2,int(n**0.5)+1):
if n % i == 0:
return False
return True
# ワーシャルフロイド (任意の2頂点の対に対して最短経路を求める)
# 計算量n^3 (nは頂点の数)
def warshall_floyd(d, n):
#d[i][j]: iからjへの最短距離
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])
return d
# ダイクストラ
def dijkstra_heap(s, edge, n):
#始点sから各頂点への最短距離
d = [10**20] * n
used = [True] * n #True:未確定
d[s] = 0
used[s] = False
edgelist = []
for a,b in edge[s]:
heapq.heappush(edgelist,a*(10**6)+b)
while len(edgelist):
minedge = heapq.heappop(edgelist)
#まだ使われてない頂点の中から最小の距離のものを探す
if not used[minedge%(10**6)]:
continue
v = minedge%(10**6)
d[v] = minedge//(10**6)
used[v] = False
for e in edge[v]:
if used[e[1]]:
heapq.heappush(edgelist,(e[0]+d[v])*(10**6)+e[1])
return d
# 素因数分解
def factorization(n):
arr = []
temp = n
for i in range(2, int(-(-n**0.5//1))+1):
if temp%i==0:
cnt=0
while temp%i==0:
cnt+=1
temp //= i
arr.append([i, cnt])
if temp!=1:
arr.append([temp, 1])
if arr==[]:
arr.append([n, 1])
return arr
# 2数の最小公倍数
def lcm(x, y):
return (x * y) // gcd(x, y)
# リストの要素の最小公倍数
def lcm_list(numbers):
return reduce(lcm, numbers, 1)
# リストの要素の最大公約数
def gcd_list(numbers):
return reduce(gcd, numbers)
# 素数判定
# limit以下の素数を列挙
def eratosthenes(limit):
A = [i for i in range(2, limit+1)]
P = []
while True:
prime = min(A)
if prime > sqrt(limit):
break
P.append(prime)
i = 0
while i < len(A):
if A[i] % prime == 0:
A.pop(i)
continue
i += 1
for a in A:
P.append(a)
return P
# 同じものを含む順列
def permutation_with_duplicates(L):
if L == []:
return [[]]
else:
ret = []
# set(集合)型で重複を削除、ソート
S = sorted(set(L))
for i in S:
data = L[:]
data.remove(i)
for j in permutation_with_duplicates(data):
ret.append([i] + j)
return ret
# ここから書き始める
r, c = map(int, input().split())
sy, sx = map(int, input().split())
sy -= 1
sx -= 1
gy, gx = map(int, input().split())
gy -= 1
gx -= 1
grid = [[j for j in input()] for i in range(r)]
# print(grid)
queue = deque([[sy, sx, 0]])
ans = 0
visited = [[0 for j in range(c)] for i in range(r)]
while queue:
y, x, cnt = queue.popleft()
if y == gy and x == gx:
ans = cnt
break
if visited[y][x] == 1:
continue
visited[y][x] = 1
for i in [[1, 0], [-1, 0], [0, 1], [0, -1]]:
if 0 <= y + i[0] < r and 0 <= x + i[1] < c:
queue.append([y + i[0], x + i[1], cnt + 1])
print(ans)