結果

問題 No.3306 Life is Easy?
ユーザー lif4635
提出日時 2025-10-05 16:13:14
言語 PyPy3
(7.3.15)
結果
TLE  
実行時間 -
コード長 6,492 bytes
コンパイル時間 273 ms
コンパイル使用メモリ 82,484 KB
実行使用メモリ 106,428 KB
最終ジャッジ日時 2025-10-05 16:13:28
合計ジャッジ時間 10,115 ms
ジャッジサーバーID
(参考情報)
judge2 / judge3
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ファイルパターン 結果
other AC * 4 TLE * 2 -- * 29
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ソースコード

diff #

# input
import sys
input = sys.stdin.readline
II = lambda : int(input())
MI = lambda : map(int, input().split())
LI = lambda : [int(a) for a in input().split()]
SI = lambda : input().rstrip()
LLI = lambda n : [[int(a) for a in input().split()] for _ in range(n)]
LSI = lambda n : [input().rstrip() for _ in range(n)]
MI_1 = lambda : map(lambda x:int(x)-1, input().split())
LI_1 = lambda : [int(a)-1 for a in input().split()]

mod = 998244353
inf = 1001001001001001001
ordalp = lambda s : ord(s)-65 if s.isupper() else ord(s)-97
ordallalp = lambda s : ord(s)-39 if s.isupper() else ord(s)-97
yes = lambda : print("Yes")
no = lambda : print("No")
yn = lambda flag : print("Yes" if flag else "No")

prinf = lambda ans : print(ans if ans < 1000001001001001001 else -1)
alplow = "abcdefghijklmnopqrstuvwxyz"
alpup = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
alpall = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
URDL = {'U':(-1,0), 'R':(0,1), 'D':(1,0), 'L':(0,-1)}
DIR_4 = [[-1,0],[0,1],[1,0],[0,-1]]
DIR_8 = [[-1,0],[-1,1],[0,1],[1,1],[1,0],[1,-1],[0,-1],[-1,-1]]
DIR_BISHOP = [[-1,1],[1,1],[1,-1],[-1,-1]]
prime60 = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59]
sys.set_int_max_str_digits(0)
# sys.setrecursionlimit(10**6)
# import pypyjit
# pypyjit.set_param('max_unroll_recursion=-1')

from collections import defaultdict,deque
from heapq import heappop,heappush
from bisect import bisect_left,bisect_right
DD = defaultdict
BSL = bisect_left
BSR = bisect_right

import heapq


class mcf_graph:
    n = 1
    pos = []
    g = [[]]

    def __init__(self, N):
        self.n = N
        self.pos = []
        self.g = [[] for i in range(N)]

    def add_edge(self, From, To, cap, cost):
        assert 0 <= From and From < self.n
        assert 0 <= To and To < self.n
        m = len(self.pos)
        self.pos.append((From, len(self.g[From])))
        self.g[From].append(
            {"to": To, "rev": len(self.g[To]), "cap": cap, "cost": cost}
        )
        self.g[To].append(
            {"to": From, "rev": len(self.g[From]) - 1, "cap": 0, "cost": -cost}
        )

    def get_edge(self, i):
        m = len(self.pos)
        assert 0 <= i and i < m
        _e = self.g[self.pos[i][0]][self.pos[i][1]]
        _re = self.g[_e["to"]][_e["rev"]]
        return {
            "from": self.pos[i][0],
            "to": _e["to"],
            "cap": _e["cap"] + _re["cap"],
            "flow": _re["cap"],
            "cost": _e["cost"],
        }

    def edges(self):
        m = len(self.pos)
        result = [{} for i in range(m)]
        for i in range(m):
            tmp = self.get_edge(i)
            result[i]["from"] = tmp["from"]
            result[i]["to"] = tmp["to"]
            result[i]["cap"] = tmp["cap"]
            result[i]["flow"] = tmp["flow"]
            result[i]["cost"] = tmp["cost"]
        return result

    def flow(self, s, t, flow_limit=-1 - (-1 << 63)):
        return self.slope(s, t, flow_limit)[-1]

    def slope(self, s, t, flow_limit=-1 - (-1 << 63)):
        assert 0 <= s and s < self.n
        assert 0 <= t and t < self.n
        assert s != t
        """
         variants (C = maxcost):
         -(n-1)C <= dual[s] <= dual[i] <= dual[t] = 0
         reduced cost (= e.cost + dual[e.from] - dual[e.to]) >= 0 for all edge
        """
        dual = [0 for i in range(self.n)]
        dist = [0 for i in range(self.n)]
        pv = [0 for i in range(self.n)]
        pe = [0 for i in range(self.n)]
        vis = [False for i in range(self.n)]

        def dual_ref():
            for i in range(self.n):
                dist[i] = -1 - (-1 << 63)
                pv[i] = -1
                pe[i] = -1
                vis[i] = False
            que = []
            heapq.heappush(que, (0, s))
            dist[s] = 0
            while que:
                v = heapq.heappop(que)[1]
                if vis[v]:
                    continue
                vis[v] = True
                if v == t:
                    break
                """
                 dist[v] = shortest(s, v) + dual[s] - dual[v]
                 dist[v] >= 0 (all reduced cost are positive)
                 dist[v] <= (n-1)C
                """
                for i in range(len(self.g[v])):
                    e = self.g[v][i]
                    if vis[e["to"]] or (not (e["cap"])):
                        continue
                    """
                     |-dual[e.to]+dual[v]| <= (n-1)C
                     cost <= C - -(n-1)C + 0 = nC
                    """
                    cost = e["cost"] - dual[e["to"]] + dual[v]
                    if dist[e["to"]] - dist[v] > cost:
                        dist[e["to"]] = dist[v] + cost
                        pv[e["to"]] = v
                        pe[e["to"]] = i
                        heapq.heappush(que, (dist[e["to"]], e["to"]))
            if not (vis[t]):
                return False
            for v in range(self.n):
                if not (vis[v]):
                    continue
                dual[v] -= dist[t] - dist[v]
            return True

        flow = 0
        cost = 0
        prev_cost = -1
        result = [(flow, cost)]
        while flow < flow_limit:
            if not (dual_ref()):
                break
            c = flow_limit - flow
            v = t
            while v != s:
                c = min(c, self.g[pv[v]][pe[v]]["cap"])
                v = pv[v]
            v = t
            while v != s:
                self.g[pv[v]][pe[v]]["cap"] -= c
                self.g[v][self.g[pv[v]][pe[v]]["rev"]]["cap"] += c
                v = pv[v]
            d = -dual[s]
            flow += c
            cost += c * d
            if prev_cost == d:
                result.pop()
            result.append((flow, cost))
            prev_cost = cost
        return result



n, m = MI()
a = LLI(n)

# 適当なマッチングにおいて
# 差を最大化したい

g = mcf_graph(2 + n + n * m + m + 10)

s = (m + 1) * n + 1
t = s + 1

off = t + 1

h = n // 2
for i in range(h): # 前半
    d = (m + 1) * i # 日付用
    g.add_edge(s, d, 1, 0)
    for j in range(m):
        p = i * (m + 1) + j + 1 # ノード
        g.add_edge(d, p, 1, a[i][j])
        g.add_edge(p, off + j, 1, 0)

inf = 10 ** 9 
for i in range(n - h, n):
    d = (m + 1) * i # 日付用
    g.add_edge(d, t, 1, 0)
    for j in range(m):
        p = i * (m + 1) + j + 1 # ノード
        g.add_edge(p, d, 1, inf - a[i][j])
        g.add_edge(off + j, p, 1, 0)

f, r = g.flow(s, t)
print(inf * h - r)
0