[AcWing 6] 多重背包问题 III

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点击查看代码

#include<iostream>
#include<cstring>

using namespace std;

const int N = 1010, M = 20010;

int n, m;
int v[N], w[N], s[N];
int f[M], g[M], q[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 ++) {
		memcpy(g, f, sizeof f);
		for (int j = 0; j < v[i]; j ++) {
			int hh = 0, tt = -1;
			for (int k = j; k <= m; k += v[i]) {
				if (hh <= tt && q[hh] < k - s[i] * v[i])	hh ++;
				while (hh <= tt && g[q[tt]] - (q[tt] - j) / v[i] * w[i] <= g[k] - (k - j) / v[i] * w[i])
					tt --;
				q[++ tt] = k;
				f[k] = g[q[hh]] + (k - q[hh]) / v[i] * w[i];
			}
		}
	}
	cout << f[m] << endl;
	return 0;
}

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1. 单调队列优化
   $f[i][j] = max(f[i-1][j],f[i-1][j-v]+w,f[i-1][j-2 \cdot v]+2 \cdot w,\cdots,f[i-1][j-s \cdot v]+s \cdot w)$
   $f[i][j-v] = max(f[i-1][j-v],f[i-1][j-2 \cdot v]+ w,\cdots,f[i-1][j-s \cdot v]+s \cdot w,f[i-1][j-3 \cdot v] + 2 \cdot w,f[i-1][j- s \cdot v] + (s-1) \cdot w ,f[i-1][j-(s+1) \cdot v] + s \cdot w)$
   $\cdots \cdots$
   观察 $f[i][j]$ 和 $f[i][j-v]$ 可以发现，除去偏移量 $w$，剩下的部分 $f[i-1][j-v], f[i-1][j-2 \cdot v], \cdots, f[i-1][j - s \cdot v]$ 是相同的，可以用单调队列来维护一段区间的最大值，将更新这一步的复杂度降低到线性