結果

問題 No.1301 Strange Graph Shortest Path
ユーザー marroncastlemarroncastle
提出日時 2020-11-29 20:19:15
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 1,291 ms / 3,000 ms
コード長 1,687 bytes
コンパイル時間 593 ms
コンパイル使用メモリ 82,388 KB
実行使用メモリ 219,000 KB
最終ジャッジ日時 2024-09-13 02:07:17
合計ジャッジ時間 38,296 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge2
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 * 2
other AC * 33
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 # 

# Dijkstra O(FElog(V))
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, 0, cost, None]
    backward = forward[4] = [fr, 0, 0, -cost, forward]
    self.G[fr].append(forward)
    self.G[to].append(backward)

  def minCostFlow(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 -1

      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
        if e[4][2]==0:
          e[2] += d
        else:
          e[4][2] -= d
        e[4][1] += d
        v = prv_v[v]
    return res

n, m = map(int, input().split())
graph = MinCostFlow(n)
for i in range(m):
  u, v, c, d = map(int, input().split())
  graph.add_edge(u-1, v-1, 1, c)
  graph.add_edge(v-1, u-1, 1, c)
  graph.add_edge(u-1, v-1, 1, d)
  graph.add_edge(v-1, u-1, 1, d)
print(graph.minCostFlow(0, n-1, 2))
0