結果

問題 No.1545 [Cherry 2nd Tune N] Anthem
ユーザー ygd.
提出日時 2021-06-15 23:15:44
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 2,400 bytes
コンパイル時間 730 ms
コンパイル使用メモリ 82,304 KB
実行使用メモリ 107,648 KB
最終ジャッジ日時 2024-12-28 02:29:30
合計ジャッジ時間 27,482 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge2
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ファイルパターン 結果
other AC * 31 WA * 35 TLE * 1
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

from heapq import heapify, heappop, heappush
from collections import deque
def main():
    N,S,T,K = map(int,input().split()); INF = float("inf")
    S-=1;T-=1 #0-index
    X = list(map(int,input().split()))
    G = [[] for _ in range(N)]
    M = int(input())
    for _ in range(M):
        A,B,Y = map(int,input().split())
        A-=1;B-=1
        G[A].append((Y,B))
    Q = deque([])
    Q.append(S)
    d_bfs = [INF]*N
    d_bfs[S] = 1
    while Q:
        v = Q.popleft()
        for cost, u in G[v]:
            if d_bfs[u] != INF: continue
            d_bfs[u] = d_bfs[v] + 1
            Q.append(v)
    if d_bfs[T] == INF:
        print("Impossible");exit()
    
    def dijkstra_heap2(s,G,t,K):
        INF = float("inf")
        #S:start, V: node, E: Edge, G: Graph
        V = len(G)
        #dp[i][j]: i番目の歌を歌って今j曲目(i番目の歌を含む)
        if K <= 10:
            MAX = 20
        elif K <= 130:
            MAX = 1000
        else:
            MAX = pow(10,5)
        dp = [[INF]*MAX for _ in range(V)]
        #d = [INF for _ in range(V)]
        dp[s][1] = X[s]
        prev = [[-1]*MAX for _ in range(V)]
        PQ = []
        heappush(PQ,(X[s],s,1)) #時間, 位置, 何曲目
        while PQ:
            c,v,n = heappop(PQ)
            if v == t and n >= K:
                break
            if dp[v][n] < c:
                continue
            dp[v][n] = c
            if n+1 >= MAX: #これ以上は配列外参照
                continue
            for cost,u in G[v]:
                if dp[u][n+1] <= cost + X[u] + dp[v][n]:
                    continue
                dp[u][n+1] = cost + X[u] + dp[v][n]
                prev[u][n+1] = v
                heappush(PQ,(dp[u][n+1], u, n+1))
        return dp, prev
    d, keiro = dijkstra_heap2(S,G,T,K)
    #print(d)
    #print(keiro)
    ans = INF
    num = INF #何曲か
    for i in range(K,len(d[T])):
        if d[T][i] < ans:
            ans = d[T][i]
            num = i
    if ans == INF:
        print("Impossible")
    else:
        print("Possible")
        print(ans)
        print(num)
        ret = [T+1] #1-index
        #num -= 1
        now = T
        while num > 1:
            pre = keiro[now][num]
            ret.append(pre+1) #1-index
            now = pre
            num -= 1
        ret.reverse()
        print(*ret)
if __name__ == '__main__':
    main()
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