結果

問題 No.1678 Coin Trade (Multiple)
ユーザー chineristAC
提出日時 2021-09-10 22:30:02
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 2,001 bytes
コンパイル時間 322 ms
コンパイル使用メモリ 82,396 KB
実行使用メモリ 106,568 KB
最終ジャッジ日時 2024-06-12 01:19:20
合計ジャッジ時間 7,381 ms
ジャッジサーバーID
(参考情報)
judge4 / judge5
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 WA * 3
other TLE * 1 -- * 55
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 #

from heapq import heappush, heappop
class MinCostFlow:
    INF = 10**18

    def __init__(self, N):
        self.N = N
        self.G = [[] for i in range(N)]

    def add_edge(self, fr, to, cap, cost):
        forward = [to, cap, cost, None]
        backward = forward[3] = [fr, 0, -cost, forward]
        self.G[fr].append(forward)
        self.G[to].append(backward)

    def flow(self, s, t, f):
        N = self.N; G = self.G
        INF = MinCostFlow.INF

        res = 0
        H = [0]*N
        prv_v = [0]*N
        prv_e = [None]*N

        d0 = [INF]*N
        dist = [INF]*N

        while f:
            dist[:] = d0
            dist[s] = 0
            que = [(0, s)]

            while que:
                c, v = heappop(que)
                if dist[v] < c:
                    continue
                r0 = dist[v] + H[v]
                for e in G[v]:
                    w, cap, cost, _ = e
                    if cap > 0 and r0 + cost - H[w] < dist[w]:
                        dist[w] = r = r0 + cost - H[w]
                        prv_v[w] = v; prv_e[w] = e
                        heappush(que, (r, w))
            if dist[t] == INF:
                return None

            for i in range(N):
                H[i] += dist[i]

            d = f; v = t
            while v != s:
                d = min(d, prv_e[v][1])
                v = prv_v[v]
            f -= d
            res += d * H[t]
            v = t
            while v != s:
                e = prv_e[v]
                e[1] -= d
                e[3][1] += d
                v = prv_v[v]
        return res

import sys,random

input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())

N,K = mi()
G = MinCostFlow(N)
A = [0 for i in range(N)]
for i in range(N):
    A[i],M = mi()
    B = li()
    for pi in B:
        print(pi-1,i,K,A[pi-1]-A[i])
        G.add_edge(pi-1,i,1,A[pi-1]-A[i])
for i in range(N-1):
    G.add_edge(i,i+1,K,0)

f = G.flow(0,N-1,K)
print(-f)
0