結果

問題 No.1324 Approximate the Matrix
コンテスト
ユーザー chineristAC
提出日時 2022-03-16 18:28:04
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 1,028 ms / 2,000 ms
コード長 2,216 bytes
コンパイル時間 734 ms
コンパイル使用メモリ 82,184 KB
実行使用メモリ 89,848 KB
最終ジャッジ日時 2024-09-24 18:58:15
合計ジャッジ時間 13,666 ms
ジャッジサーバーID
（参考情報） 		judge1 / 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 AC * 3
other AC * 42
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
                else:
                    pass
                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))
                    else:
                        pass
            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

N,K = map(int,input().split())
A = list(map(int,input().split()))
B = list(map(int,input().split()))
p = [list(map(int,input().split())) for i in range(N)]

G = MinCostFlow(2*N+2)
for i in range(1,N+1):
    G.add_edge(0,i,A[i-1],0)
    for j in range(N+1,2*N+1):
        for k in range(min(A[i-1],B[j-N-1])):
            G.add_edge(i,j,1,2*k+1-2*p[i-1][j-N-1])
for j in range(N+1,2*N+1):
    G.add_edge(j,2*N+1,B[j-N-1],0)

f = G.flow(0,2*N+1,K)
for i in range(N):
    for j in range(N):
        f += p[i][j]**2

print(f)
0