[AcWing 1019] 庆功会

多重背包朴素做法 时间复杂度 $O(n \cdot m \cdot s)$

总体时间复杂度 $500 \times 6000 \times 10 = 3 \times 10^{7}$

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点击查看代码

#include<iostream>

using namespace std;

const int N = 510, M = 6010;

int n, m;
int v[N], w[N], s[N];
int f[N][M];

int main()
{
	cin >> n >> m;
	for (int i = 1; i <= n; i ++)	cin >> v[i] >> w[i] >> s[i];
	for (int i = 1; i <= n; i ++)
		for (int j = 0; j <= m; j ++)
			for (int k = 0; k <= s[i] && k * v[i] <= j; k ++)
				f[i][j] = max(f[i][j], f[i - 1][j - k * v[i]] + k * w[i]);
	cout << f[n][m] << endl;
	return 0;
}

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多重背包二进制优化 时间复杂度 $O(n \cdot m \cdot log(s))$

总体时间复杂度 $500 \times 6000 \times log(10) \approx 1 \times 10^{7}$

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点击查看代码

#include<iostream>

using namespace std;

const int N = 1e5 + 10;

int n, m;
int v[N], w[N], cnt;
int f[N];

int main()
{
	cin >> n >> m;
	for (int i = 0; i < n; i ++) {
		int a, b, s;
		cin >> a >> b >> s;
		int k = 1;
		while (k <= s) {
			cnt ++;
			v[cnt] = a * k;
			w[cnt] = b * k;
			s -= k;
			k *= 2;
		}
		if (s) {
			cnt ++;
			v[cnt] = a * s;
			w[cnt] = b * s;
		}
	}
	for (int i = 1; i <= cnt; i ++)
		for (int j = m; j >= v[i]; j --)
			f[j] = max(f[j], f[j - v[i]] + w[i]);
	cout << f[m] << endl;
	return 0;
}

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