東京大学プログラミングコンテスト2011: H. キャッシュ戦略
- http://www.utpc.jp/2011/problems/cache.html
- https://beta.atcoder.jp/contests/utpc2011/tasks/utpc2011_8
- http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=2266
solution
最小費用流。$O(N^2 M \sum w_j)$。
editorialにある図を見ると早いので大きく略すが、次に注意:
- 辺の張り方について、左端はひとつずれる
- 連続して同じクエリが来る場合は事前に潰しておかねばならない
implementation
#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; (i) < int(n); ++ (i))
#define REP_R(i, n) for (int i = int(n) - 1; (i) >= 0; -- (i))
#define ALL(x) begin(x), end(x)
using namespace std;
template <class T> using reversed_priority_queue = priority_queue<T, vector<T>, greater<T> >;
template <class T> inline void chmin(T & a, T const & b) { a = min(a, b); }
template <class T>
struct edge { int to; T cap, cost; int rev; };
template <class T>
void add_edge(vector<vector<edge<T> > > & graph, int from, int to, T cap, T cost) {
graph[from].push_back((edge<T>) { to, cap, cost, int(graph[ to].size()) });
graph[ to].push_back((edge<T>) { from, 0, - cost, int(graph[from].size()) - 1 });
}
/**
* @brief minimum-cost flow with primal-dual method
* @note mainly O(V^2UC) for U is the sum of capacities and C is the sum of costs. and additional O(VE) if negative edges exist
*/
template <class T>
T min_cost_flow_destructive(int src, int dst, T flow, vector<vector<edge<T> > > & graph) {
T result = 0;
vector<T> potential(graph.size());
if (0 < flow) { // initialize potential when negative edges exist (slow). you can remove this if unnecessary
fill(ALL(potential), numeric_limits<T>::max());
potential[src] = 0;
while (true) { // Bellman-Ford algorithm
bool updated = false;
REP (e_from, graph.size()) for (auto & e : graph[e_from]) if (e.cap) {
if (potential[e_from] == numeric_limits<T>::max()) continue;
if (potential[e.to] > potential[e_from] + e.cost) {
potential[e.to] = potential[e_from] + e.cost; // min
updated = true;
}
}
if (not updated) break;
}
}
while (0 < flow) {
// update potential using dijkstra
vector<T> distance(graph.size(), numeric_limits<T>::max()); // minimum distance
vector<int> prev_v(graph.size()); // constitute a single-linked-list represents the flow-path
vector<int> prev_e(graph.size());
{ // initialize distance and prev_{v,e}
reversed_priority_queue<pair<T, int> > que; // distance * vertex
distance[src] = 0;
que.emplace(0, src);
while (not que.empty()) { // Dijkstra's algorithm
T d; int v; tie(d, v) = que.top(); que.pop();
if (potential[v] == numeric_limits<T>::max()) continue; // for unreachable nodes
if (distance[v] < d) continue;
// look round the vertex
REP (e_index, graph[v].size()) {
// consider updating
edge<T> e = graph[v][e_index];
int w = e.to;
if (potential[w] == numeric_limits<T>::max()) continue;
T d1 = distance[v] + e.cost + potential[v] - potential[w]; // updated distance
if (0 < e.cap and d1 < distance[e.to]) {
distance[w] = d1;
prev_v[w] = v;
prev_e[w] = e_index;
que.emplace(d1, w);
}
}
}
}
if (distance[dst] == numeric_limits<T>::max()) return -1; // no such flow
REP (v, graph.size()) {
if (potential[v] == numeric_limits<T>::max()) continue;
potential[v] += distance[v];
}
// finish updating the potential
// let flow on the src->dst minimum path
T delta = flow; // capacity of the path
for (int v = dst; v != src; v = prev_v[v]) {
chmin(delta, graph[prev_v[v]][prev_e[v]].cap);
}
flow -= delta;
result += delta * potential[dst];
for (int v = dst; v != src; v = prev_v[v]) {
edge<T> & e = graph[prev_v[v]][prev_e[v]]; // reference
e.cap -= delta;
graph[v][e.rev].cap += delta;
}
}
return result;
}
constexpr int inf = 1e9 + 7;
int main() {
// input
int m, n, k; scanf("%d%d%d", &m, &n, &k);
vector<int> w(n); REP (i, n) scanf("%d", &w[i]);
vector<int> a(k);
REP (i, k) {
scanf("%d", &a[i]);
-- a[i];
}
// solve
a.erase(unique(ALL(a)), a.end()); // remove consecutive same elements
k = a.size();
int base = 0;
vector<vector<edge<int> > > g(k);
vector<int> last(n, -1);
REP_R (i, k) {
base += w[a[i]];
if (i < k - 1) {
add_edge(g, i, i + 1, inf, 0);
if (last[a[i]] != -1) {
add_edge(g, i + 1, last[a[i]], 1, - w[a[i]]);
}
}
last[a[i]] = i;
}
int result = base + min_cost_flow_destructive(0, k - 1, m - 1, g);
// output
printf("%d\n", result);
return 0;
}