結果
| 問題 | No.1678 Coin Trade (Multiple) |
| ユーザー |
👑  tatyam
|
| 提出日時 | 2021-09-10 23:08:39 |
| 言語 | C++23 (gcc 12.3.0 + boost 1.83.0) |
| 結果 |
TLE
|
| 実行時間 | - |
| コード長 | 11,989 bytes |
| コンパイル時間 | 3,612 ms |
| コンパイル使用メモリ | 258,988 KB |
| 実行使用メモリ | 11,196 KB |
| 最終ジャッジ日時 | 2023-09-02 21:10:44 |
| 合計ジャッジ時間 | 28,332 ms |
|
ジャッジサーバーID （参考情報） |
judge12 / judge15 |
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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 | 1 ms
4,372 KB |
| testcase_01 | AC | 2 ms
4,372 KB |
| testcase_02 | AC | 2 ms
4,372 KB |
| testcase_03 | AC | 169 ms
6,308 KB |
| testcase_04 | AC | 534 ms
10,220 KB |
| testcase_05 | AC | 622 ms
10,284 KB |
| testcase_06 | AC | 446 ms
11,196 KB |
| testcase_07 | AC | 330 ms
6,280 KB |
| testcase_08 | AC | 357 ms
8,716 KB |
| testcase_09 | AC | 589 ms
10,088 KB |
| testcase_10 | AC | 27 ms
4,564 KB |
| testcase_11 | AC | 443 ms
9,788 KB |
| testcase_12 | AC | 30 ms
4,656 KB |
| testcase_13 | AC | 696 ms
10,144 KB |
| testcase_14 | AC | 218 ms
6,004 KB |
| testcase_15 | AC | 440 ms
9,844 KB |
| testcase_16 | AC | 267 ms
9,512 KB |
| testcase_17 | AC | 598 ms
10,204 KB |
| testcase_18 | AC | 1 ms
4,372 KB |
| testcase_19 | AC | 1 ms
4,368 KB |
| testcase_20 | AC | 2 ms
4,372 KB |
| testcase_21 | AC | 2 ms
4,368 KB |
| testcase_22 | AC | 2 ms
4,372 KB |
| testcase_23 | AC | 1 ms
4,372 KB |
| testcase_24 | AC | 2 ms
4,372 KB |
| testcase_25 | AC | 1 ms
4,372 KB |
| testcase_26 | AC | 2 ms
4,368 KB |
| testcase_27 | AC | 2 ms
4,368 KB |
| testcase_28 | AC | 1 ms
4,372 KB |
| testcase_29 | AC | 1 ms
4,368 KB |
| testcase_30 | AC | 2 ms
4,372 KB |
| testcase_31 | AC | 2 ms
4,368 KB |
| testcase_32 | AC | 1 ms
4,372 KB |
| testcase_33 | AC | 2 ms
4,368 KB |
| testcase_34 | AC | 2 ms
4,368 KB |
| testcase_35 | AC | 2 ms
4,368 KB |
| testcase_36 | AC | 2 ms
4,368 KB |
| testcase_37 | AC | 2 ms
4,368 KB |
| testcase_38 | AC | 1 ms
4,368 KB |
| testcase_39 | AC | 1 ms
4,372 KB |
| testcase_40 | AC | 1 ms
4,372 KB |
| testcase_41 | AC | 2 ms
4,368 KB |
| testcase_42 | AC | 2 ms
4,368 KB |
| testcase_43 | AC | 2 ms
4,368 KB |
| testcase_44 | AC | 2 ms
4,368 KB |
| testcase_45 | AC | 1 ms
4,376 KB |
| testcase_46 | AC | 2 ms
4,368 KB |
| testcase_47 | AC | 2 ms
4,368 KB |
| testcase_48 | AC | 899 ms
10,716 KB |
| testcase_49 | AC | 936 ms
10,884 KB |
| testcase_50 | AC | 949 ms
10,364 KB |
| testcase_51 | AC | 869 ms
10,532 KB |
| testcase_52 | AC | 892 ms
10,484 KB |
| testcase_53 | AC | 893 ms
10,364 KB |
| testcase_54 | AC | 893 ms
10,588 KB |
| testcase_55 | AC | 866 ms
10,688 KB |
| testcase_56 | AC | 1,028 ms
10,632 KB |
| testcase_57 | AC | 893 ms
10,552 KB |
| testcase_58 | TLE | - |
ソースコード
#include <bits/stdc++.h>
using namespace std;
// https://judge.yosupo.jp/submission/57330
struct linked_lists {
int L, N;
vector<int> next, prev;
// L: lists are [0...L), N: elements are [0,...N)
explicit linked_lists(int L = 0, int N = 0) { assign(L, N); }
int rep(int l) const { return l + N; } // "representative" of list l
int head(int l) const { return next[rep(l)]; }
int tail(int l) const { return prev[rep(l)]; }
bool empty(int l) const { return next[rep(l)] == rep(l); }
void push_front(int l, int n) { link(rep(l), n, head(l)); }
void push_back(int l, int n) { link(tail(l), n, rep(l)); }
void insert_before(int i, int n) { link(prev[i], n, i); }
void insert_after(int i, int n) { link(i, n, next[i]); }
void erase(int n) { link(prev[n], next[n]); }
void pop_front(int l) { link(rep(l), next[head(l)]); }
void pop_back(int l) { link(prev[tail(l)], rep(l)); }
void clear() {
iota(begin(next) + N, end(next), N); // sets next[rep(l)] = rep(l)
iota(begin(prev) + N, end(prev), N); // sets prev[rep(l)] = rep(l)
}
void assign(int L, int N) {
this->L = L, this->N = N;
next.resize(N + L), prev.resize(N + L), clear();
}
private:
inline void link(int u, int v) { next[u] = v, prev[v] = u; }
inline void link(int u, int v, int w) { link(u, v), link(v, w); }
};
// Iterate through elements (call them i) of the list #l in "lists"
#define FOR_EACH_IN_LINKED_LIST(i, l, lists) \
for (int z##i = l, i = lists.head(z##i); i != lists.rep(z##i); i = lists.next[i])
/**
* Network simplex for minimum cost circulation with fixed supply/demand at nodes.
* Supports negative costs, negative cost cycles, self-loops and multiple edges fine.
*
* Flow should be large enough to hold sum of supplies and edge flows
* Cost should be large enough to hold sum of absolute costs * 2V (usually >=64 bits)
*
* Complexity: O(V) expected pivot, O(E) worst-case
*
* Usage:
* network_simplex<int, long> mcc(V);
* for (edges...) {
* mcc.add(u, v, cap, cost);
* }
* for (nodes...) {
* mcc.set_supply(u, supply);
* mcc.set_demand(u, demand);
* }
* bool feasible = mcc.mincost_circulation();
* auto mincost = mcc.circulation_cost();
*
* References:
* LEMON network_simplex.h
* OCW MIT MIT15_082JF10_av16.pdf
* OCW MIT MIT15_082JF10_lec16.pdf
*/
template <typename Flow, typename Cost>
struct network_simplex {
// we number the vertices 0,...,V-1, R is given number V.
explicit network_simplex(int V) : V(V), node(V + 1) {}
void add(int u, int v, Flow cap, Cost cost) {
assert(0 <= u && u < V && 0 <= v && v < V);
edge.push_back({{u, v}, cap, cost}), E++;
}
void set_supply(int u, Flow supply) { node[u].supply = supply; }
void set_demand(int u, Flow demand) { node[u].supply = -demand; }
auto get_supply(int u) const { return node[u].supply; }
auto get_potential(int u) const { return node[u].pi; }
auto get_flow(int e) const { return edge[e].flow; }
auto get_cost(int e) const { return edge[e].cost; }
auto reduced_cost(int e) const {
auto [u, v] = edge[e].node;
return edge[e].cost + node[u].pi - node[v].pi;
}
template <typename CostSum = Cost>
auto circulation_cost() const {
CostSum sum = 0;
for (int e = 0; e < E; e++) {
sum += edge[e].flow * CostSum(edge[e].cost);
}
return sum;
}
void verify_spanning_tree() const {
for (int e = 0; e < E; e++) {
assert(0 <= edge[e].flow && edge[e].flow <= edge[e].cap);
assert(edge[e].flow == 0 || reduced_cost(e) <= 0);
assert(edge[e].flow == edge[e].cap || reduced_cost(e) >= 0);
}
}
bool mincost_circulation() {
static constexpr bool INFEASIBLE = false, OPTIMAL = true;
// Check trivialities: positive cap[e] and sum of supplies is 0
Flow sum_supply = 0;
for (int u = 0; u < V; u++) {
sum_supply += node[u].supply;
}
if (sum_supply != 0) {
return INFEASIBLE;
}
for (int e = 0; e < E; e++) {
if (edge[e].cap < 0) {
return INFEASIBLE;
}
}
// Compute inf_cost as sum of all costs + 1, and reset the flow network
Cost inf_cost = 1;
for (int e = 0; e < E; e++) {
edge[e].flow = 0;
edge[e].state = STATE_LOWER;
inf_cost += abs(edge[e].cost);
}
edge.resize(E + V); // make space for V artificial edges
bfs.resize(V + 1);
children.assign(V + 1, V + 1);
// Add V artificial edges with infinite cost and initial supply for feasible flow
int root = V;
node[root] = {-1, -1, 0, 0};
for (int u = 0, e = E; u < V; u++, e++) {
// spanning tree links
node[u].parent = root, node[u].pred = e;
children.push_back(root, u);
auto supply = node[u].supply;
if (supply >= 0) {
node[u].pi = -inf_cost;
edge[e] = {{u, root}, supply, inf_cost, supply, STATE_TREE};
} else {
node[u].pi = inf_cost;
edge[e] = {{root, u}, -supply, inf_cost, -supply, STATE_TREE};
}
}
// We want to, hopefully, find a pivot edge in O(sqrt(E)). This can be tuned.
block_size = max(int(ceil(sqrt(E + V))), min(10, V + 1));
next_arc = 0;
// Pivot pivot pivot
int in_arc = select_pivot_edge();
while (in_arc != -1) {
pivot(in_arc);
in_arc = select_pivot_edge();
}
// If there is >0 flow through an artificial edge, the problem is infeasible.
for (int e = E; e < E + V; e++) {
if (edge[e].flow > 0) {
edge.resize(E);
return INFEASIBLE;
}
}
edge.resize(E);
return OPTIMAL;
}
private:
enum ArcState : int8_t { STATE_UPPER = -1, STATE_TREE = 0, STATE_LOWER = 1 };
struct Node {
int parent, pred;
Flow supply;
Cost pi;
};
struct Edge {
array<int, 2> node; // [0]->[1]
Flow cap;
Cost cost;
Flow flow = 0;
ArcState state = STATE_LOWER;
};
int V, E = 0, next_arc = 0, block_size = 0;
vector<Node> node;
vector<Edge> edge;
linked_lists children;
vector<int> bfs; // scratchpad for downwards bfs and evert
int select_pivot_edge() {
// block search: check block_size edges looping, and pick the lowest reduced cost
Cost minimum = 0;
int in_arc = -1;
int count = block_size, seen_edges = E + V;
for (int& e = next_arc; seen_edges-- > 0; e = e + 1 == E + V ? 0 : e + 1) {
if (minimum > edge[e].state * reduced_cost(e)) {
minimum = edge[e].state * reduced_cost(e);
in_arc = e;
}
if (--count == 0 && minimum < 0) {
break;
} else if (count == 0) {
count = block_size;
}
}
return in_arc;
}
void pivot(int in_arc) {
// Find lca of u_in and v_in with two pointers technique
auto [u_in, v_in] = edge[in_arc].node;
int a = u_in, b = v_in;
while (a != b) {
a = node[a].parent == -1 ? v_in : node[a].parent;
b = node[b].parent == -1 ? u_in : node[b].parent;
}
int lca = a;
// Orient the edge so that we add flow along u_in->v_in
if (edge[in_arc].state == STATE_UPPER) {
swap(u_in, v_in);
}
// Let's find the saturing flow push
enum OutArcSide { SAME_EDGE, U_IN_SIDE, V_IN_SIDE };
OutArcSide side = SAME_EDGE;
Flow flow_delta = edge[in_arc].cap; // how much we can push to saturate
int u_out = -1; // the exiting arc
// Go up from u_in to lca, break ties by prefering lower vertices
for (int u = u_in; u != lca && flow_delta > 0; u = node[u].parent) {
int e = node[u].pred;
bool edge_down = u == edge[e].node[1];
Flow to_saturate = edge_down ? edge[e].cap - edge[e].flow : edge[e].flow;
if (flow_delta > to_saturate) {
flow_delta = to_saturate;
u_out = u;
side = U_IN_SIDE;
}
}
// Go up from v_in to lca, break ties by prefering higher vertices
for (int u = v_in; u != lca; u = node[u].parent) {
int e = node[u].pred;
bool edge_up = u == edge[e].node[0];
Flow to_saturate = edge_up ? edge[e].cap - edge[e].flow : edge[e].flow;
if (flow_delta >= to_saturate) {
flow_delta = to_saturate;
u_out = u;
side = V_IN_SIDE;
}
}
// Augment along the cycle if we can push anything
if (flow_delta > 0) {
auto delta = edge[in_arc].state * flow_delta;
edge[in_arc].flow += delta;
for (int u = edge[in_arc].node[0]; u != lca; u = node[u].parent) {
int e = node[u].pred;
edge[e].flow += u == edge[e].node[0] ? -delta : delta;
}
for (int u = edge[in_arc].node[1]; u != lca; u = node[u].parent) {
int e = node[u].pred;
edge[e].flow += u == edge[e].node[0] ? delta : -delta;
}
}
// Return now if we didn't change the spanning tree. The state of in_arc flipped.
if (side == SAME_EDGE) {
edge[in_arc].state = ArcState(-edge[in_arc].state);
return;
}
// Basis exchange: Replace out_arc with in_arc in the spanning tree
int out_arc = node[u_out].pred;
edge[in_arc].state = STATE_TREE;
edge[out_arc].state = edge[out_arc].flow ? STATE_UPPER : STATE_LOWER;
// Put u_in on the same side as u_out
if (side == V_IN_SIDE) {
swap(u_in, v_in);
}
// Evert: Walk up from u_in to u_out, and fix parent/pred/child pointers downwards
int i = 0, S = 0;
for (int u = u_in; u != u_out; u = node[u].parent) {
bfs[S++] = u;
}
assert(S <= V);
for (i = S - 1; i >= 0; i--) {
int u = bfs[i], p = node[u].parent;
children.erase(p); // remove p from its children list and add it to u's
children.push_back(u, p);
node[p].parent = u;
node[p].pred = node[u].pred;
}
children.erase(u_in); // remove u_in from its children list and add it to v_in's
children.push_back(v_in, u_in);
node[u_in].parent = v_in;
node[u_in].pred = in_arc;
// Fix potentials: Visit the subtree of u_in (pi_delta is not 0).
Cost current_pi = reduced_cost(in_arc);
Cost pi_delta = u_in == edge[in_arc].node[0] ? -current_pi : current_pi;
bfs[0] = u_in;
for (i = 0, S = 1; i < S; i++) {
int u = bfs[i];
node[u].pi += pi_delta;
FOR_EACH_IN_LINKED_LIST (v, u, children) { bfs[S++] = v; }
}
}
};
using ll = long long;
int main(){
cin.tie(nullptr);
ios::sync_with_stdio(false);
ll N, K;
cin >> N >> K;
network_simplex<ll, ll> g(N);
g.set_supply(0, K);
g.set_demand(N - 1, K);
for(ll i = 0; i < N - 1; i++) g.add(i, i + 1, K, 1LL << 30);
vector<ll> A(N);
for(ll i = 0; i < N; i++){
ll M;
cin >> A[i] >> M;
while(M--){
ll B;
cin >> B;
B--;
if(A[B] < A[i]) g.add(B, i, 1, ((i - B) << 30) - (A[i] - A[B]));
}
}
g.mincost_circulation();
cout << ((K * (N - 1)) << 30) - g.circulation_cost() << endl;
}