結果

問題 No.1502 Many Simple Additions
ユーザー mkawa2
提出日時 2024-01-02 02:09:16
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 2,772 bytes
コンパイル時間 284 ms
コンパイル使用メモリ 82,504 KB
実行使用メモリ 116,588 KB
最終ジャッジ日時 2024-09-27 17:38:29
合計ジャッジ時間 7,143 ms
ジャッジサーバーID
(参考情報)
judge1 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 5
other AC * 21 WA * 18
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

import sys
# sys.setrecursionlimit(1000005)
# sys.set_int_max_str_digits(200005)
int1 = lambda x: int(x)-1
pDB = lambda *x: print(*x, end="\n", file=sys.stderr)
p2D = lambda x: print(*x, sep="\n", end="\n\n", file=sys.stderr)
def II(): return int(sys.stdin.readline())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def LI1(): return list(map(int1, sys.stdin.readline().split()))
def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
def SI(): return sys.stdin.readline().rstrip()
# dij = [(0, 1), (-1, 0), (0, -1), (1, 0)]
dij = [(0, 1), (-1, 0), (0, -1), (1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)]
# inf = -1-(-1 << 31)
inf = -1-(-1 << 62)
md = 10**9+7
# md = 998244353
from collections import defaultdict
n,m,k=LI()
to=[[] for _ in range(n)]
uv2z=defaultdict(int)
for _ in range(m):
u,v,z=LI()
u,v=u-1,v-1
to[u].append(v)
to[v].append(u)
uv2z[u,v]=uv2z[v,u]=z
def dfs(root=0):
aa[root]=1
stack = [(u, root) for u in to[root]]
lmin=rmin=inf
lmax=rmax=-inf
while stack:
u, p = stack.pop()
z = uv2z[u, p]
if aa[u]:
if aa[u]==aa[p]:
# 2x=au(z-bu-bp)
x2=aa[u]*(z-bb[u]-bb[p])
if x2 < 2 or x2 > k*2 or x2&1:
print(0)
exit()
x=x2//2
if xx[root]==None:
xx[root]=x
elif xx[root]!=x:
print(0)
exit()
elif bb[u]!=bb[p]:
print(0)
exit()
else:
parent[u]=p
aa[u]=-aa[p]
bb[u]=z-bb[p]
if aa[u]==1:
lmin,lmax=min(lmin,bb[u]),max(lmax,bb[u])
else:
rmin, rmax = min(rmin, bb[u]), max(rmax, bb[u])
for v in to[u]:
if aa[v]:continue
stack.append((v,u))
# x+lmin>=1
# x>=1-lmin
# -x+rmin>=1
# x<=rmin-1
# x+lmax<=k
# x<=k-lmax
# -x+rmax<=k
# x>=rmax-k
if xx[root]==None:
mn=max(1-lmin,rmax-k,1)
mx=min(rmin-1,k-lmax,k)
a1=max(0,mx-mn+1)
mn=max(1-lmin,rmax-k+1,1)
mx=min(rmin-1,k-1-lmax,k-1)
a2=max(0,mx-mn+1)
return a1,a2
x=xx[root]
if x+lmin<1 or -x+rmin<1:
print(0)
exit()
a1=a2=0
if x+lmax<=k and -x+rmax<=k:a1=1
if x+lmax<k and -x+rmax<k:a2=1
return a1,a2
parent = [-1]*n
aa=[0]*n
bb=[0]*n
xx=[None]*n
s1=s2=1
for u in range(n):
if aa[u]:continue
a1,a2=dfs(u)
s1*=a1
s2*=a2
s1%=md
s2%=md
# print(u,a1,a2,s1,s2)
# print(aa)
# print(bb)
# print(xx)
print((s1-s2)%md)
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