結果

問題 No.1812 Uribo Road
ユーザー 👑 SPD_9X2
提出日時 2022-01-14 22:57:04
言語 PyPy3
(7.3.15)
結果
RE  
実行時間 -
コード長 2,423 bytes
コンパイル時間 194 ms
コンパイル使用メモリ 82,436 KB
実行使用メモリ 99,336 KB
最終ジャッジ日時 2024-11-20 13:33:59
合計ジャッジ時間 9,511 ms
ジャッジサーバーID
（参考情報） 		judge3 / 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 * 4
other AC * 24 RE * 6
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ソースコード

diff #
プレゼンテーションモードにする

"""
bitdpっぽくやるのね…
10^7...
厳しめ
dp[bit][最後][LorR] = mincost
"""
from sys import stdin
import heapq
def Dijkstra(lis,start):
    ret = [float("inf")] * len(lis)
    ret[start] = 0
    q = [(0,start)]
    while len(q) > 0:
        ncost,now = heapq.heappop(q)
        if ncost != ret[now]:
            continue
        for nex,ecost in lis[now]:
            if ret[nex] > ncost + ecost:
                ret[nex] = ncost + ecost
                heapq.heappush(q , (ret[nex] , nex))
    return ret
N,M,K = map(int,stdin.readline().split())
R = list(map(int,stdin.readline().split()))
for i in range(K):
    R[i] -= 1
lis = [ [] for i in range(M) ]
ABC = []
for i in range(M):
    A,B,C = map(int,stdin.readline().split())
    A -= 1
    B -= 1
    ABC.append( (A,B,C) )
    lis[A].append( (B,C) )
    lis[B].append( (A,C) )
ddic = {}
ddic[0] = Dijkstra(lis,0)
for nr in R:
    l,r,_ = ABC[nr]
    if l not in ddic:
        ddic[l] = Dijkstra(lis,l)
    if r not in ddic:
        ddic[r] = Dijkstra(lis,r)
#print (ddic)
inf = float("inf")
dp = [ [[inf,inf] for j in range(K)] for i in range(2**K) ]
# dp[bit][最後][LorR] = mincost
for i,nr in enumerate(R):
    l,r,c = ABC[nr]
    dp[2**i][i][0] = ddic[0][r] + c
    dp[2**i][i][1] = ddic[0][l] + c
def popcnt(x):
    ret = 0
    while x:
        ret += x%2
        x //= 2
    return ret
for bit in range(2**K):
    for lastNR in range(K):
        for lastSide in range(2):
            lastL,lastR,_ = ABC[R[lastNR]]
            lastV = [lastL,lastR][lastSide]
            
            if dp[bit][lastNR][lastSide] == inf:
                continue
            for newNR in range(K):
                newbit = 2**newNR | bit
                if bit == newbit:
                    continue
                newL,newR,newC = ABC[R[newNR]]
                
                dp[newbit][newNR][0] = min(dp[newbit][newNR][0] ,
                                           dp[bit][lastNR][lastSide] + ddic[lastV][newR] + newC )
                dp[newbit][newNR][1] = min(dp[newbit][newNR][1] ,
                                           dp[bit][lastNR][lastSide] + ddic[lastV][newL] + newC )
ans = inf
bit = 2**K-1
for lastNR in range(K):
    for lastSide in range(2):
        lastL,lastR,_ = ABC[R[lastNR]]
        lastV = [lastL,lastR][lastSide]
        ans = min(ans , dp[bit][lastNR][lastSide] + ddic[lastV][N-1])
print (ans)
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