/
subset_sum.hpp
205 lines (193 loc) · 4.88 KB
/
subset_sum.hpp
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#include "ds/my_bitset.hpp"
#include "enumerate/bits.hpp"
// O(N MAX(vals))
template <typename T>
vc<int> subset_sum_solution_1(vc<T>& vals, int target) {
int n = len(vals);
if (n == 0) return {};
int mx = MAX(vals);
int b = 0, sb = 0;
while (b < n && sb + vals[b] <= target) { sb += vals[b++]; }
if (b == n && sb != target) return {};
int off = target - mx + 1;
vc<int> dp(2 * mx, -1);
vv(int, PAR, n, 2 * mx, -1);
dp[sb - off] = b;
FOR3(i, b, n) {
auto newdp = dp;
auto& par = PAR[i];
int a = vals[i];
FOR(j, mx) {
if (chmax(newdp[j + a], dp[j])) { par[j + a] = -2; }
}
FOR3_R(j, mx, 2 * mx) {
FOR3_R(k, max(dp[j], 0), newdp[j]) {
if (chmax(newdp[j - vals[k]], k)) par[j - vals[k]] = k;
}
}
swap(dp, newdp);
}
if (dp[mx - 1] == -1) return {};
vc<bool> use(n);
int i = n - 1, j = mx - 1;
while (i >= b) {
int p = PAR[i][j];
if (p == -2) {
use[i] = !use[i];
j -= vals[i--];
}
elif (p == -1) { --i; }
else {
use[p] = !use[p];
j += vals[p];
}
}
while (i >= 0) {
use[i] = !use[i];
--i;
}
vc<int> I;
FOR(i, n) if (use[i]) I.eb(i);
ll sm = 0;
for (auto&& i: I) sm += vals[i];
assert(sm == target);
return I;
}
// O(N target / w)
template <typename T>
vc<int> subset_sum_solution_2(vc<T>& vals, int target) {
int n = len(vals);
auto I = argsort(vals);
My_Bitset dp(1, 1);
vc<int> last(target + 1, -1);
FOR(k, n) {
int v = vals[I[k]];
if (v > target) continue;
My_Bitset newdp = dp;
int new_size = len(dp) + v;
newdp.resize(new_size);
newdp.or_to_range(v, new_size, dp);
if (len(newdp) > target + 1) newdp.resize(target + 1);
// update したところをメモ
FOR(i, len(newdp.dat)) {
u64 upd = (i < len(dp.dat) ? dp.dat[i] : u64(0)) ^ newdp.dat[i];
enumerate_bits_64(upd, [&](int p) -> void { last[(i << 6) | p] = I[k]; });
}
swap(dp, newdp);
}
if (target >= len(dp) || !dp[target]) return {};
vc<int> ANS;
while (target > 0) {
int i = last[target];
ANS.eb(i);
target -= vals[i];
}
return ANS;
}
// O(sum^{1.5} / w)
// sum=10^6 で 150ms:https://codeforces.com/contest/755/problem/F
template <typename T>
vc<int> subset_sum_solution_3(vc<T>& vals, int target) {
int SM = SUM<int>(vals);
int N = len(vals);
vvc<int> IDS(SM + 1);
FOR(i, N) IDS[vals[i]].eb(i);
vc<pair<int, int>> par(N, {-1, -1});
vc<int> grp_vals;
vc<int> raw_idx;
FOR(x, 1, SM + 1) {
auto& I = IDS[x];
while (len(I) >= 3) {
int a = POP(I), b = POP(I);
int c = len(par);
par.eb(a, b);
IDS[2 * x].eb(c);
}
for (auto& i: I) {
grp_vals.eb(x);
raw_idx.eb(i);
}
}
auto I = subset_sum_solution_2<int>(grp_vals, target);
vc<int> ANS;
for (auto& i: I) {
vc<int> st = {raw_idx[i]};
while (len(st)) {
auto c = POP(st);
if (c < N) {
ANS.eb(c);
continue;
}
auto [a, b] = par[c];
st.eb(a), st.eb(b);
}
}
return ANS;
}
template <typename T>
vc<int> subset_sum_solution_4(vc<T>& vals, T target) {
if (target <= 0) return {};
int N = len(vals);
int M = N / 2;
auto calc = [&](int L, int R) -> vc<T> {
int n = R - L;
vc<T> dp = {0};
FOR(i, n) {
T a = vals[L + i];
vc<T> dp1(len(dp));
FOR(k, len(dp)) dp1[k] = dp[k] + a;
vc<T> newdp;
merge(all(dp), all(dp1), back_inserter(newdp));
swap(dp, newdp);
}
return dp;
};
auto restore = [&](int L, int R, T v) -> vc<int> {
int n = R - L;
vc<T> dp(1 << n);
FOR(i, n) FOR(s, 1 << i) dp[s | 1 << i] = dp[s] + vals[L + i];
FOR(s, 1 << n) {
if (dp[s] == v) {
vc<int> I;
FOR(i, n) if (s >> i & 1) I.eb(L + i);
return I;
}
}
assert(0);
return {};
};
auto dp1 = calc(0, M);
auto dp2 = calc(M, N);
int t = 0;
FOR_R(s, len(dp1)) {
while (t + 1 < len(dp2) && dp1[s] + dp2[t + 1] <= target) { ++t; }
if (dp1[s] + dp2[t] == target) {
vc<int> I1 = restore(0, M, dp1[s]);
vc<int> I2 = restore(M, N, dp2[t]);
I1.insert(I1.end(), all(I2));
return I1;
}
}
return {};
}
template <typename T>
vc<int> subset_sum(vc<T>& vals, T target) {
if (target <= 0) return {};
int n = len(vals);
if (n == 0) return {};
int mx = MAX(vals);
// しきい値の調整をしていない
// solution 1: O(N mx))
// solution 2: O(N target / w)
// solution 3: O(sum^1.5 / w)
// solution 4: O(2^(N/2))
double x1 = double(len(vals)) * mx;
double x2 = double(len(vals)) * target / 500.0;
double x3 = pow(SUM<double>(vals), 1.5) / 500.0;
double x4 = pow(2.0, 0.5 * len(vals));
double mi = min({x1, x2, x3, x4});
if (x1 == mi) return subset_sum_solution_1(vals, target);
if (x2 == mi) return subset_sum_solution_2(vals, target);
if (x3 == mi) return subset_sum_solution_3(vals, target);
return subset_sum_solution_4(vals, target);
}