結果

問題 No.1324 Approximate the Matrix
ユーザー chineristAC
提出日時 2020-12-22 12:36:44
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 1,268 ms / 2,000 ms
コード長 2,100 bytes
コンパイル時間 286 ms
コンパイル使用メモリ 82,192 KB
実行使用メモリ 90,488 KB
最終ジャッジ日時 2024-09-21 14:06:22
合計ジャッジ時間 17,778 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge5
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ファイルパターン 結果
sample AC * 3
other AC * 42
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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 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
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(A[i-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)
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