// this solves subtask 1, 3, 4, 5. the complexity is alright,
// but for larger n we get MLE. instead of binary lifting, we
// should probably use some other base, HLD, those linear jump pointers..
#include "dungeons.h"
#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 50001; // subtasks 1, 3, 4, 5
const int MAX_LOG = 25; // 2^24 > 10^7
// dp[i][j][k] = we're at i, simulate for 2^j steps and
// win against an opponent if strength < 2^k (we don't gain strength)
struct {
int end; // where do we end up at?
long long gained; // total strength gained on the path
// max(sum(gained) - opponent) along the simulation,
// for all opponents we lose against
long long max;
} dp[MAX_N][MAX_LOG][MAX_LOG];
int n, s[MAX_N], w[MAX_N];
// because p[i] = s[i], every time we lose, our strength doubles
struct Subtask2 {
struct {
int end;
long long gained;
long long min; // min(sum(gained) - opponent)
} dp[400001][MAX_LOG];
int l[400001];
Subtask2(vector<int>& l_) {
copy_n(l_.begin(), n, l);
l[n] = n;
for (int j = 0; j < MAX_LOG; j++)
for (int i = 0; i <= n; i++) {
auto& x = dp[i][j];
if (j == 0)
if (i == n) {
x.end = n;
x.gained = x.min = 0;
} else {
x.end = w[i];
x.gained = s[i];
x.min = -s[i];
}
else {
auto& a = dp[i][j - 1];
auto& b = dp[a.end][j - 1];
x.end = b.end;
x.gained = a.gained + b.gained;
x.min = min(a.min, a.gained + b.min);
}
}
}
long long simulate(int at, int s_init) {
long long strength = 0;
auto lose = [&](int i, int j) {
return strength + dp[i][j].min < 0;
};
while (at != n) {
// jump until we lose
for (int j = MAX_LOG - 1; j >= 0; j--)
if (!lose(at, j)) {
auto& x = dp[at][j];
at = x.end;
strength += x.gained;
}
// lose (if we're at the end, this adds 0)
strength += s[at];
at = l[at];
}
return strength;
}
};
optional<Subtask2> sub2; // don't waste memory :)
void init(int n_, vector<int> s_, vector<int> p,
vector<int> w_, vector<int> l) {
n = n_;
copy_n(s_.begin(), n, s);
copy_n(w_.begin(), n, w);
w[n] = n;
if (n > 50000) {
sub2.emplace(l);
return;
}
for (int j = 0; j < MAX_LOG; j++)
for (int i = 0; i <= n; i++)
for (int k = 0; k < MAX_LOG; k++) {
auto& x = dp[i][j][k];
if (j == 0)
if (i == n) {
x.end = n;
x.gained = x.max = 0;
} else if (s[i] < (1 << k)) { // we win
x.end = w[i];
x.gained = s[i];
x.max = LLONG_MIN;
} else {
x.end = l[i];
x.gained = p[i];
// 1 step behind (we didn't gain 'gained' yet)
x.max = -s[i];
}
else {
auto& a = dp[i][j - 1][k];
auto& b = dp[a.end][j - 1][k];
x.end = b.end;
x.gained = a.gained + b.gained;
x.max = max(a.max, a.gained + b.max);
}
}
}
long long simulate(int at, int s_init) {
if (n > 50000)
return sub2->simulate(at, s_init);
long long strength = s_init;
int k = 0;
auto advance_k = [&] { // msb
if (strength >= (1 << 24))
k = 24;
else
while ((1 << (k + 1)) <= strength)
k++;
};
// find first time our dp takes the wrong turn. once this happens,
// the msb will change => happens only log(something) times
// at this vertex we win:
// strength + gained - opponent >= 0
// => strength + dp.max >= 0
// but dp loses (dp.max is only calculated for losing edges)
auto wrong_turn = [&](int i, int j, int k) {
return strength + dp[i][j][k].max >= 0;
};
while (at != n) {
advance_k();
// jump to point where msb changes
for (int j = MAX_LOG - 1; j >= 0; j--)
if (!wrong_turn(at, j, k)) { // doesn't change => jump forward
auto& x = dp[at][j][k];
at = x.end;
strength += x.gained;
} // else it changes, so don't jump forward
// now wrong_turn(at, 0, k) holds, so we win
strength += s[at];
at = w[at];
}
return strength;
}
