[AcWing 1013] 机器分配

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点击查看代码

#include<iostream>

using namespace std;

const int N = 20;

int n, m;
int w[N][N];
int f[N][N];
int way[N];

int main()
{
	cin >> n >> m;
	for (int i = 1; i <= n; i ++)
		for (int j = 1; j <= m; j ++)
			cin >> w[i][j];
	for (int i = 1; i <= n; i ++)
		for (int j = 0; j <= m; j ++) {
			f[i][j] = f[i - 1][j];
			for (int k = 1; k <= j; k ++)
				f[i][j] = max(f[i][j], f[i - 1][j - k] + w[i][k]);
		}
	cout << f[n][m] << endl;
	int j = m;
	for (int i = n; i >= 0; i --)
		for (int k = 0; k <= j; k ++)
			if (f[i][j] == f[i - 1][j - k] + w[i][k]) {
				way[i] = k;
				j -= k;
				break;
			}
	for (int i = 1; i <= n; i ++)	cout << i << ' ' << way[i] << endl;
	return 0;
}

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1. 状态表示
   $f[i][j]$ 表示从前 $i$ 组选，且总体积不超过 $j$ 的选法的最大值
2. 状态计算
   $f[i][j] = max(f[i - 1][j], f[i - 1][j - k] + w[i][k])$
   ( $k$ 表示从第 $i$ 组中选择的物品个数)