結果

問題 No.2604 Initial Motion
ユーザー pitP
提出日時 2024-01-13 01:02:11
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 873 ms / 3,000 ms
コード長 8,392 bytes
コンパイル時間 275 ms
コンパイル使用メモリ 82,304 KB
実行使用メモリ 83,568 KB
最終ジャッジ日時 2024-09-28 00:55:13
合計ジャッジ時間 17,657 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
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ファイルパターン 結果
sample AC * 3
other AC * 39
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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

# https://yukicoder.me/submissions/943504
# https://atcoder.jp/contests/practice2/submissions/20654283
import heapq
class mcf_graph_int_cost:
    """It solves Minimum-cost flow problem.
    
    """
    def __init__(self, n):
        """It creates a directed graph with n vertices and 0 edges.
        Cap and Cost are the type of the capacity and the cost, respectively.
        Constraints
        -----------
        >   0 <= n <= 10 ** 8
        >   Cap and Cost are int.
        Complexity
        ----------
        >   O(n)
        """
        self.n = n
        self._edges = []
    def add_edge(self, from_, to, cap, cost):
        """It adds an edge oriented from `from_` to `to` with capacity `cap` and cost `cost`. 
        It returns an integer k such that this is the k-th edge that is added.
        Constraints
        -----------
        >   0 <= from_, to < n
        >   0 <= cap, cost
        Complexity
        ----------
        >   O(1) amortized
        """
        # assert 0 <= from_ < self.n
        # assert 0 <= to < self.n
        # assert 0 <= cap
        # assert 0 <= cost
        m = len(self._edges)
        self._edges.append(self.__class__.edge(from_, to, cap, 0, cost))
        return m
    class edge:
        def __init__(self, from_, to, cap, flow, cost):
            self.from_ = from_
            self.to = to
            self.cap = cap
            self.flow = flow
            self.cost = cost
    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.
        Constraints
        -----------
        >   0 <= i < m
        Complexity
        ----------
        >   O(1)
        """
        # assert 0 <= i < len(self._edges)
        return self._edges[i]
    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.
        """
        return self._edges.copy()
    def _dual_ref(self, s, t):
        self.dist = [4611686018427387903] * self.n
        self.vis = [False] * self.n
        self.que_min.clear()
        self.que.clear()
        que_push_que = []
        self.dist[s] = 0
        self.que_min.append(s)
        while self.que_min or self.que or que_push_que:
            if self.que_min:
                v = self.que_min.pop()
            else:
                while que_push_que:
                    heapq.heappush(self.que, que_push_que.pop())
                v = heapq.heappop(self.que) & 9223372036854775807
            if self.vis[v]:
                continue
            self.vis[v] = True
            if v == t:
                break
            dual_v = self.dual[v]
            dist_v = self.dist[v]
            for i in range(self.start[v], self.start[v + 1]):
                e = self.elist[i]
                if not e.cap:
                    continue
                cost = e.cost - self.dual[e.to] + dual_v
                if self.dist[e.to] - dist_v > cost:
                    dist_to = dist_v + cost
                    self.dist[e.to] = dist_to
                    self.prev_e[e.to] = e.rev
                    if dist_to == dist_v:
                        self.que_min.append(e.to)
                    else:
                        que_push_que.append((dist_to << 64) + e.to)
        if not self.vis[t]:
            return False
        for v in range(self.n):
            if not self.vis[v]:
                continue
            self.dual[v] -= self.dist[t] - self.dist[v]
        return True
    def _csr(self):
        m = len(self._edges)
        self.edge_idx = [0] * m
        redge_idx = [0] * m
        degree = [0] * self.n
        edges = []
        for i, e in enumerate(self._edges):
            self.edge_idx[i] = degree[e.from_]
            degree[e.from_] += 1
            redge_idx[i] = degree[e.to]
            degree[e.to] += 1
            edges.append((e.from_, self.__class__._edge(
                e.to, -1, e.cap - e.flow, e.cost)))
            edges.append((e.to, self.__class__._edge(
                e.from_, -1, e.flow, -e.cost)))
        self.start = [0] * (self.n + 1)
        self.elist = [0] * len(edges)
        for v, w in edges:
            self.start[v + 1] += 1
        for i in range(1, self.n + 1):
            self.start[i] += self.start[i-1]
        counter = self.start.copy()
        for v, w in edges:
            self.elist[counter[v]] = w
            counter[v] += 1
        for i, e in enumerate(self._edges):
            self.edge_idx[i] += self.start[e.from_]
            redge_idx[i] += self.start[e.to]
            self.elist[self.edge_idx[i]].rev = redge_idx[i]
            self.elist[redge_idx[i]].rev = self.edge_idx[i]
    def slope(self, s, t, flow_limit=4611686018427387903):
        """Let g be a function such that g(x) is the cost of the minimum cost s-t flow 
        when the amount of the flow is exactly x. 
        g is known to be piecewise linear.
        It returns g as the list of the changepoints, that satisfies the followings.
        >   The first element of the list is (0, 0).
        >   Both of element[0] and element[1] are strictly increasing.
        >   No three changepoints are on the same line.
        >   (1) The last element of the list is (x, g(x)), where x is the maximum amount of the s-t flow.
        >   (2) The last element of the list is (y, g(y)), where y = min(x, flow_limit).
        Constraints
        -----------
        Let x be the maximum cost among all edges.
        >   s != t
        >   You can't call min_cost_slope or min_cost_max_flow multiple times.
        >   0 <= nx <= 2 * 10 ** 9 + 1000
        Complexity
        ----------
        >   O(F (n + m) log (n + m)), where F is the amount of the flow and m is the number of added edges.
        """
        # assert 0 <= s < self.n
        # assert 0 <= t < self.n
        # assert s != t
        self._csr()
        self.dual = [0] * self.n
        self.dist = [4611686018427387903] * self.n
        self.prev_e = [0] * self.n
        self.vis = [False] * self.n
        flow = 0
        cost = 0
        prev_cost_per_flow = -1
        result = [(0, 0)]
        self.que = []
        self.que_min = []
        while flow < flow_limit:
            if not self._dual_ref(s, t):
                break
            c = flow_limit - flow
            v = t
            while v != s:
                c = min(c, self.elist[self.elist[self.prev_e[v]].rev].cap)
                v = self.elist[self.prev_e[v]].to
            v = t
            while v != s:
                e = self.elist[self.prev_e[v]]
                e.cap += c
                self.elist[e.rev].cap -= c
                v = self.elist[self.prev_e[v]].to
            d = -self.dual[s]
            flow += c
            cost += c * d
            if prev_cost_per_flow == d:
                result.pop()
            result.append((flow, cost))
            prev_cost_per_flow = d
        for i in range(len(self._edges)):
            e = self.elist[self.edge_idx[i]]
            self._edges[i].flow = self._edges[i].cap - e.cap
        return result
    def flow(self, s, t, flow_limit=4611686018427387903):
        """It augments the flow from s to t as much as possible. 
        It returns the amount of the flow and the cost.
        (1) It augments the s-t flow as much as possible.
        (2) It augments the s-t flow as much as possible, until reaching the amount of flow_limit. 
        Constraints
        -----------
        >   same as mcf_graph.slope
        Complexity
        ----------
        >   same as mcf_graph.slope
        """
        return self.slope(s, t, flow_limit)[-1]
    class _edge:
        def __init__(self, to, rev, cap, cost):
            self.to = to
            self.rev = rev
            self.cap = cap
            self.cost = cost
from collections import Counter
k,n,m = map(int,input().split())
a = Counter(map(int,input().split()))
b = list(map(int,input().split()))
g = mcf_graph_int_cost(n+2)
for i,v in a.items(): g.add_edge(0,i,v,0)
for i,v in enumerate(b,1): g.add_edge(i,n+1,v,0)
for _ in range(m):
    u,v,d = map(int,input().split())
    g.add_edge(u,v,2*k,d); g.add_edge(v,u,2*k,d)
print(g.flow(0,n+1)[1])
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