結果

問題 No.3306 Life is Easy?
ユーザー kidodesu
提出日時 2025-10-05 15:51:07
言語 PyPy3
(7.3.15)
結果
TLE  
実行時間 -
コード長 2,137 bytes
コンパイル時間 241 ms
コンパイル使用メモリ 82,296 KB
実行使用メモリ 179,604 KB
最終ジャッジ日時 2025-10-05 15:51:31
合計ジャッジ時間 6,713 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge5
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ファイルパターン 結果
other TLE * 1 -- * 34
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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

n, m = map(int, input().split())
A = []
for i in range(n):
    l = list(map(int, input().split()))
    if n % 2 and i == n // 2:
        continue
    A.append(l[:])

n -= n % 2

mf = MinCostFlow(n+2)
M = 1 << 40
s = n
t = s + 1
k = n // 2
for now in range(k):
    mf.add_edge(s, now, 1, 0)
    mf.add_edge(now+k, t, 1, 0)
    for nxt in range(k, 2*k):
        d = 0
        for i in range(m):
            d = max(d, A[nxt][i] - A[now][i])
        #print(now, nxt, d)
        mf.add_edge(now, nxt, 1, M-d)

ans = mf.flow(s, t, k)
print(M*k-ans)
0