結果
| 問題 | No.1690 Power Grid |
| コンテスト | |
| ユーザー |
ansain
|
| 提出日時 | 2021-06-30 20:25:27 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
AC
|
| 実行時間 | 867 ms / 3,000 ms |
| コード長 | 2,329 bytes |
| 記録 | |
| コンパイル時間 | 231 ms |
| コンパイル使用メモリ | 81,792 KB |
| 実行使用メモリ | 81,152 KB |
| 最終ジャッジ日時 | 2024-07-05 09:49:08 |
| 合計ジャッジ時間 | 11,289 ms |
|
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 25 |
ソースコード
import sys
from collections import defaultdict, Counter, deque
from itertools import permutations, combinations, product, combinations_with_replacement, groupby, accumulate
import operator
from math import sqrt, gcd, factorial
# from math import isqrt, prod,comb # python3.8用(notpypy)
#from bisect import bisect_left,bisect_right
#from functools import lru_cache,reduce
#from heapq import heappush,heappop,heapify,heappushpop,heapreplace
#import numpy as np
#import networkx as nx
#from networkx.utils import UnionFind
#from numba import njit, b1, i1, i4, i8, f8
#from scipy.sparse import csr_matrix
#from scipy.sparse.csgraph import shortest_path, floyd_warshall, dijkstra, bellman_ford, johnson, NegativeCycleError
# numba例 @njit(i1(i4[:], i8[:, :]),cache=True) 引数i4配列、i8 2次元配列,戻り値i1
def input(): return sys.stdin.readline().rstrip()
def divceil(n, k): return 1+(n-1)//k # n/kの切り上げを返す
def yn(hantei, yes='Yes', no='No'): print(yes if hantei else no)
def main():
mod = 10**9+7
mod2 = 998244353
n, m, k = map(int, input().split())
A = list(map(int, input().split()))
if k==1:
print(min(A))
return
cost = [[float("inf")]*n for i in range(n)]
for i in range(m):
x, y, z = map(int, input().split())
x -= 1
y -= 1
cost[x][y] = z
cost[y][x] = z
for i in range(n):
cost[i][i] = 0
for kk in range(n):
for ii in range(n):
for jj in range(n):
cost[ii][jj] = min(cost[ii][jj], cost[ii][kk] + cost[kk][jj])
ans = 10**12
dp = [10**12]*2**n
dp[0] == 0
popcnt = [0]*2**n
for bit in range(1, 2**n):
popcnt[bit] = popcnt[bit ^ (bit & (-bit))]+1
if 2 <= popcnt[bit] <= k:
for i in range(n):
if bit & 2**i:
for j in range(i+1, n):
if bit & 2**j:
dp[bit] = min(
A[i]+cost[i][j]+dp[bit ^ (2**i)], A[j]+cost[i][j]+dp[bit ^ (2**j)], dp[bit])
if popcnt[bit]==k and ans > dp[bit]:
ans=dp[bit]
elif popcnt[bit]==1:
for i in range(n):
if bit & 2**i:
dp[bit]=A[i]
print(ans)
if __name__ == '__main__':
main()
ansain