經典多重背包:n 容量,m 類物品,每類有體積 c[i]、價值 v[i]、數量 b[i]。求最大價值。
將數量 b 拆分為 1,2,4,...,2^k,b-2^k+1 份,每份視為 01 背包物品。
代碼片段:
for (int i=1; i<=m; ++i) {
int k = 1, cnt = b[i];
while (k < cnt) {
V.push_back(k * v[i]); W.push_back(k * c[i]);
cnt -= k; k <<= 1;
}
if (cnt) { V.push_back(cnt * v[i]); W.push_back(cnt * c[i]); }
}
// 01 背包
for (int i=0; i=W[i]; --j)
dp[j] = max(dp[j], dp[j-W[i]] + V[i]);
對每個餘數類 r = 0..c[i]-1,DP 狀態 dp[j] = max_{k} (dp[j-k*c] + k*v)。用單調隊列維護滑動窗口最大值,線性處理每類。
#include
#include
#include
using namespace std;
int main() {
int T; cin >> T;
while (T--) {
int n, m; cin >> n >> m;
vector V, W;
for (int i=0, c,v,b; i> c >> v >> b;
for (int k=1; k<=b; k<<=1) {
V.push_back(k*v); W.push_back(k*c);
b -= k;
}
if (b) { V.push_back(b*v); W.push_back(b*c); }
}
vector dp(n+1, 0);
for (int i=0; i=W[i]; --j)
dp[j] = max(dp[j], dp[j-W[i]] + V[i]);
cout << dp[n] << endl;
}
}