[AcWing 12] 背包问题求具体方案

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点击查看代码

#include<iostream>
#include<vector>

using namespace std;

const int N = 1010;

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

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

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1. 状态计算的时候从后往前做，\(f[1][m]\) 是最大值
2. 求字典序最小的方案时，从前往后做，只要能选，就输出