結果

問題 No.1900 Don't be Powers of 2
ユーザー とりゐとりゐ
提出日時 2022-04-08 23:12:27
言語 PyPy3
(7.3.15)
結果
TLE  
実行時間 -
コード長 6,677 bytes
コンパイル時間 471 ms
コンパイル使用メモリ 87,296 KB
実行使用メモリ 362,336 KB
最終ジャッジ日時 2023-08-19 01:48:39
合計ジャッジ時間 33,519 ms
ジャッジサーバーID
(参考情報)
judge14 / judge9
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 61 ms
71,428 KB
testcase_01 AC 62 ms
71,484 KB
testcase_02 AC 61 ms
71,484 KB
testcase_03 AC 1,181 ms
79,148 KB
testcase_04 AC 1,180 ms
79,232 KB
testcase_05 AC 1,185 ms
79,376 KB
testcase_06 AC 1,179 ms
79,116 KB
testcase_07 AC 1,184 ms
79,044 KB
testcase_08 AC 368 ms
80,992 KB
testcase_09 AC 227 ms
79,724 KB
testcase_10 AC 126 ms
78,224 KB
testcase_11 AC 168 ms
78,968 KB
testcase_12 AC 478 ms
82,460 KB
testcase_13 AC 85 ms
77,384 KB
testcase_14 AC 827 ms
78,960 KB
testcase_15 AC 86 ms
77,480 KB
testcase_16 AC 88 ms
77,272 KB
testcase_17 AC 1,094 ms
79,192 KB
testcase_18 AC 876 ms
78,804 KB
testcase_19 AC 1,181 ms
79,044 KB
testcase_20 AC 838 ms
78,800 KB
testcase_21 AC 783 ms
78,828 KB
testcase_22 AC 752 ms
78,620 KB
testcase_23 AC 321 ms
78,632 KB
testcase_24 AC 474 ms
78,612 KB
testcase_25 AC 600 ms
78,800 KB
testcase_26 AC 1,987 ms
359,456 KB
testcase_27 AC 750 ms
121,892 KB
testcase_28 AC 613 ms
84,512 KB
testcase_29 AC 582 ms
84,012 KB
testcase_30 AC 126 ms
78,588 KB
testcase_31 AC 60 ms
71,528 KB
testcase_32 AC 59 ms
71,420 KB
testcase_33 AC 1,885 ms
362,336 KB
testcase_34 AC 1,911 ms
362,020 KB
testcase_35 TLE -
testcase_36 AC 1,591 ms
299,492 KB
testcase_37 AC 793 ms
83,956 KB
testcase_38 AC 555 ms
156,340 KB
testcase_39 AC 405 ms
109,576 KB
testcase_40 AC 1,183 ms
81,048 KB
testcase_41 AC 1,168 ms
83,932 KB
testcase_42 AC 61 ms
71,140 KB
testcase_43 AC 61 ms
71,420 KB
testcase_44 AC 61 ms
71,720 KB
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ソースコード

diff #

class mf_graph:
    """It solves maximum flow problem.
    """
 
    def __init__(self, n):
        """It creates a graph of n vertices and 0 edges.
 
        Constraints
        -----------
 
        >   0 <= n <= 10 ** 8
 
        Complexity
        ----------
 
        >   O(n)
        """
        self.n = n
        self.g = [[] for _ in range(self.n)]
        self.pos = []
 
    def add_edge(self, from_, to, cap):
        """It adds an edge oriented from the vertex `from_` to the vertex `to` 
        with the capacity `cap` and the flow amount 0. 
        It returns an integer k such that this is the k-th edge that is added.
 
        Constraints
        -----------
 
        >   0 <= from_, to < n
 
        >   0 <= cap
 
        Complexity
        ----------
 
        >   O(1) amortized
        """
        # assert 0 <= from_ < self.n
        # assert 0 <= to < self.n
        # assert 0 <= cap
        m = len(self.pos)
        self.pos.append((from_, len(self.g[from_])))
        self.g[from_].append(self.__class__._edge(to, len(self.g[to]), cap))
        self.g[to].append(self.__class__._edge(from_, len(self.g[from_]) - 1, 0))
        return m
 
    class edge:
        def __init__(self, from_, to, cap, flow):
            self.from_ = from_
            self.to = to
            self.cap = cap
            self.flow = flow
 
    def get_edge(self, i):
        """It returns the current internal state of the edges.
        The edges are ordered in the same order as added by add_edge.
 
        Complexity
        ----------
 
        > O(1)
        """
        # assert 0 <= i < len(self.pos)
        _e = self.g[self.pos[i][0]][self.pos[i][1]]
        _re = self.g[_e.to][_e.rev]
        return self.__class__.edge(self.pos[i][0], _e.to, _e.cap + _re.cap, _re.cap)
 
    def edges(self):
        """It returns the current internal state of the edges.
        The edges are ordered in the same order as added by add_edge.
 
        Complexity
        ----------
 
        >   O(m), where m is the number of added edges.
        """
        result = []
        for i in range(len(self.pos)):
            _e = self.g[self.pos[i][0]][self.pos[i][1]]
            _re = self.g[_e.to][_e.rev]
            result.append(self.__class__.edge(self.pos[i][0], _e.to, _e.cap + _re.cap, _re.cap))
        return result
 
    def change_edge(self, i, new_cap, new_flow):
        """It changes the capacity and the flow amount of the ii-th edge to new_cap and new_flow, respectively. It doesn't change the capacity or the flow amount of other edges. See Appendix for further details.
 
        Constraints
        -----------
 
        >   0 <= newflow <= newcap
 
        Complexity
        ----------
 
        >   O(1)
        """
        # assert 0 <= i < len(self.pos)
        # assert 0 <= new_flow <= new_cap
        _e = self.g[self.pos[i][0]][self.pos[i][1]]
        _re = self.g[_e.to][_e.rev]
        _e.cap = new_cap - new_flow
        _re.cap = new_flow
 
    def _bfs(self, s, t):
        self.level = [-1] * self.n
        self.level[s] = 0
        q = [s]
        while q:
            nq = []
            for v in q:
                for e in self.g[v]:
                    if e.cap and self.level[e.to] == -1:
                        self.level[e.to] = self.level[v] + 1
                        if e.to == t:
                            return True
                        nq.append(e.to)
            q = nq
        return False
 
    def _dfs(self, s, t, up):
        st = [t]
        while st:
            v = st[-1]
            if v == s:
                st.pop()
                flow = up
                for w in st:
                    e = self.g[w][self.it[w]]
                    flow = min(flow, self.g[e.to][e.rev].cap)
                for w in st:
                    e = self.g[w][self.it[w]]
                    e.cap += flow
                    self.g[e.to][e.rev].cap -= flow
                return flow
            while self.it[v] < len(self.g[v]):
                e = self.g[v][self.it[v]]
                w = e.to
                cap = self.g[e.to][e.rev].cap
                if cap and self.level[v] > self.level[w]:
                    st.append(w)
                    break
                self.it[v] += 1
            else:
                st.pop()
                self.level[v] = self.n
        return 0
 
    def flow(self, s, t, flow_limit=float('inf')):
        """It augments the flow from s to t as much as possible. 
        It returns the amount of the flow augmented.
        You may call it multiple times. 
        See Appendix in the document of AC Library for further details.
 
        Constraints
        -----------
 
        >   s != t
 
        Complexity
        ----------
 
        >   O(min(n^(2/3)m, m^(3/2))) (if all the capacities are 1) or
 
        >   O(n^2 m) (general),
 
        where m is the number of added edges.
        """
        # assert 0 <= s < self.n
        # assert 0 <= t < self.n
        # assert s != t
        flow = 0
        while flow < flow_limit and self._bfs(s, t):
            self.it = [0] * self.n
            while flow < flow_limit:
                f = self._dfs(s, t, flow_limit - flow)
                if not f:
                    break
                flow += f
        return flow
 
    def min_cut(self, s):
        """It returns a vector of length n, 
        such that the i-th element is true if and only if there is a directed path from s to i in the residual network. 
        The returned vector corresponds to a s−t minimum cut after calling flow(s, t) exactly once without flow_limit. 
        See Appendix in the document of AC Library for further details.
 
        Complexity
        ----------
 
        >   O(n + m), where m is the number of added edges.
        """
        visited = [False] * self.n
        q = [s]
        while q:
            nq = []
            for p in q:
                visited[p] = True
                for e in self.g[p]:
                    if e.cap and not visited[e.to]:
                        nq.append(e.to)
            q = nq
        return visited
 
    class _edge:
        def __init__(self, to, rev, cap):
            self.to = to
            self.rev = rev
            self.cap = cap
def popcount(m):
  return bin(m).count('1')
n=int(input())
a=list(map(int,input().split()))

s,t=n,n+1
g=mf_graph(n+2)
inf=min(n,10)

for i in range(n):
  if popcount(a[i])%2==0:
    g.add_edge(s,i,1)
  else:
    g.add_edge(i,t,1)

for i in range(n):
  for j in range(i+1,n):
    if popcount(a[i]^a[j])==1:
      if popcount(a[i])%2==0:
        g.add_edge(i,j,inf)
      else:
        g.add_edge(j,i,inf)
print(n-g.flow(s,t))
0