求背包容量恰好被装满时,最大价值为多少?
需要将数组设置为很小的负数,这样表示状态不可达
设置f[0] = 0表示背包容量为0是物体价值为0,初始化
#include <algorithm> #include <cstring> #include <iostream> using namespace std; const int N = 1010; int v[N], w[N], f[N], g[N]; int main() { int n, m; cin >> n >> m; for (int i = 1; i <= n; i ++) { cin >> v[i] >> w[i]; } for (int i = 1; i <= n; i ++) { for (int j = m; j >= v[i]; j --) { f[j] = max(f[j], f[j - v[i]] + w[i]); } } cout << f[m] << endl; memset(f, -0x3f, sizeof f); f[0] = 0; for (int i = 1; i <= n; i ++) { for (int j = m; j >= v[i]; j --) { if(f[j - v[i]] != 0x3f) f[j] = max(f[j], f[j - v[i]] + w[i]); } } if(f[m] < 0) f[m] = 0; cout << f[m]; return 0; }