結果
| 問題 |
No.1324 Approximate the Matrix
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2022-04-14 10:48:36 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 2,704 bytes |
| コンパイル時間 | 176 ms |
| コンパイル使用メモリ | 82,304 KB |
| 実行使用メモリ | 89,984 KB |
| 最終ジャッジ日時 | 2024-12-24 05:54:59 |
| 合計ジャッジ時間 | 12,337 ms |
|
ジャッジサーバーID (参考情報) |
judge1 / judge4 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 41 WA * 1 |
ソースコード
#その2 最初はコストが全部正の場合に使えるダイクストラ
from heapq import heappush,heappop
class MinCostFlow:
inf = 10 ** 10
def __init__(self,N):
self.N = N
self.G = [[] for _ in range(N)]
self.H = [0] * N
self.edge = []
def add_edge(self,fr,to,cap,cost):
e = [to,cap,cost,None]
r = e[3] = [fr,0,-cost,e]
self.G[fr].append(e)
self.G[to].append(r)
self.edge.append(e)
def get_edge(self,i):
return self.edge[i]
def flow(self,s,t,f):
N = self.N
G = self.G
inf = MinCostFlow.inf
res = 0
H = self.H #ポテンシャル、コストが最初は正なので、最初は0でいい
prev_v = [0] * N
prev_e = [None] * N
d0 = [inf] * N
dist = [inf] * N
while f:
dist[:] = d0
dist[s] = 0
q = [(0,s)]
while q:
c,v = heappop(q)
if dist[v] < c:continue
if v == t:break
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]
prev_v[w] = v
prev_e[w] = e
heappush(q,(r,w))
if dist[t] == inf:
return None
"""
for i in range(N):
H[i] += dist[i]
"""
for i in range(N):
if dist[i] < inf:
H[i] += dist[i]
d = f
v = t
while v != s:
d = min(d,prev_e[v][1])
v = prev_v[v]
f -= d
res += d * H[t]
v = t
while v != s:
e = prev_e[v]
e[1] -= d
e[3][1] += d
v = prev_v[v]
return res
import sys
rr = sys.stdin
N,K = map(int,rr.readline().split())
A = list(map(int,rr.readline().split()))
B = list(map(int,rr.readline().split()))
P = [list(map(int,rr.readline().split())) for _ in range(N)]
mincost = MinCostFlow(N + N + 2)
s = N + N
t = N + N + 1
inf = 400
for i in range(N):
for j in range(N):
v = N + j
for k in range(1,min(A[i],B[j])+1):
mincost.add_edge(i,v,1,inf + (2 * (k - 1) + 1 - 2 * P[i][j]))
for i in range(N):
mincost.add_edge(s,i,A[i],0)
for j in range(N):
mincost.add_edge(j + N,t,B[j],0)
ans = 0
for i in range(N):
for j in range(N):
ans += P[i][j] ** 2
ans += mincost.flow(s,t,K) - inf * K
print(ans)