Hard
Maximize Value of Function in a Ball Passing Game — C++
Full explanation · Time O(n) · Space O(n)
// Time: O(n)
// Space: O(n)
// graph, prefix sum, two pointers, sliding window
class Solution {
public:
long long getMaxFunctionValue(vector<int>& receiver, long long k) {
const auto& find_cycles = [](const auto& adj) {
vector<pair<int, int>> result;
vector<int> lookup(size(adj));
for (int i = 0, idx = 0; i < size(adj); ++i) {
int u = i, prev = idx;
while (!lookup[u]) {
lookup[u] = ++idx;
u = adj[u];
}
if (lookup[u] > prev) {
result.emplace_back(u, idx - lookup[u] + 1);
}
}
return result;
};
const auto& cycles = find_cycles(receiver);
vector<pair<int, int>> lookup(size(receiver), {-1, -1});
vector<vector<int64_t>> prefixes(size(cycles), vector<int64_t>(1));
const auto& find_prefixes = [&](const auto& cycles) {
for (int idx = 0; idx < size(cycles); ++idx) {
auto [u, l] = cycles[idx];
for (int i = 0; i < l; ++i) {
lookup[u] = {idx, i};
prefixes[idx].emplace_back(prefixes[idx][i] + u);
u = receiver[u];
}
}
};
const auto& get_sum = [](const auto& prefix, int64_t i, int64_t cnt) {
const int64_t l = size(prefix) - 1;
const int64_t q = cnt / l;
const int64_t r = cnt % l;
return (q * prefix.back() +
(prefix[min(i + r, l)] - prefix[i]) +
(prefix[max((i + r) - l, static_cast<int64_t>(0))] - prefix[0]));
};
const auto& start_inside_cycle = [&]() {
int64_t result = 0;
for (auto [u, l] : cycles) {
for (int _ = 0; _ < l; ++_) {
const auto& [idx, i] = lookup[u];
result = max(result, get_sum(prefixes[idx], i, k + 1));
u = receiver[u];
}
}
return result;
};
const auto& start_outside_cycle = [&]() {
int64_t result = 0;
vector<int> degree(size(receiver));
for (const auto& x : receiver) {
++degree[x];
}
for (int u = 0; u < size(receiver); ++u) {
if (degree[u]) {
continue;
}
int64_t curr = 0;
deque<int> dq;
int v = u;
while (lookup[v].first == -1) {
curr += v;
dq.emplace_back(v);
if (size(dq) == k + 1) {
result = max(result, curr);
curr -= dq.front(); dq.pop_front();
}
v = receiver[v];
}
const auto& [idx, i] = lookup[v];
while (!empty(dq)) {
result = max(result, curr + get_sum(prefixes[idx], i, (k + 1) - size(dq)));
curr -= dq.front(); dq.pop_front();
}
}
return result;
};
find_prefixes(cycles);
return max(start_inside_cycle(), start_outside_cycle());
}
};
// Time: O(nlogk)
// Space: O(nlogk)
// binary lifting
class Solution2 {
public:
long long getMaxFunctionValue(vector<int>& receiver, long long k) {
const auto& bit_length = [](int64_t x) {
return x != 0 ? 8 * sizeof(x) - __builtin_clzll(x) : 1;
};
const int l = bit_length(k + 1);
vector<vector<int>> P(l, vector<int>(size(receiver)));
P[0] = receiver;
vector<vector<int64_t>> S(l, vector<int64_t>(size(receiver)));
iota(begin(S[0]), end(S[0]), 0);
for (int i = 1; i < l; ++i) {
for (int u = 0; u < size(receiver); ++u) {
P[i][u] = P[i - 1][P[i - 1][u]];
S[i][u] = S[i - 1][u] + S[i - 1][P[i - 1][u]];
}
}
int64_t result = 0;
for (int u = 0; u < size(receiver); ++u) {
int64_t curr = 0;
for (int i = 0, v = u; i < l; ++i) {
if ((k + 1) & (1ll << i)) {
curr += S[i][v];
v = P[i][v];
}
}
result = max(result, curr);
}
return result;
}
};