题目描述
有 N 种物品和一个容量是 V 的背包,每种物品都有无限件可用
算法1
(暴力枚举) $O(n^2)$
三层循环
C++ 代码
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010;
int f[N][N];
int v[N], w[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 = 0; j <= m; j++)
for (int k = 0; 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;
}
算法2
优化为两层循环
通过观察f[i][j]展开,发现f[i][j] 与 f[i, j - v]的关系,把k干掉了
代码
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010;
int f[N][N];
int v[N], w[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 = 0; j <= m; j++)
{
f[i][j] = f[i - 1][j];
if (j >= v[i]) f[i][j] = max(f[i][j], f[i][j - v[i]] + w[i]);
}
cout << f[n][m] << endl;
return 0;
}
算法3
优化到一维
这里的j层循环,不用从大到小逆序推,因为算法2里,求的就是f[i][j - v[i]]
这样正序推,f[j - v[i]]算的刚好是第i层的f[j - v[i]]
和01背包是不同的,01背包要求得是i-1层的f[j - v[i]]
代码
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010;
int f[N];
int v[N], w[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 = v[i]; j <= m; j++)
f[j] = max(f[j], f[j - v[i]] + w[i]);
cout << f[m] << endl;
return 0;
}
算法1f[i][j] = max(f[i][j], f[i - 1][j - k * v[i]] + k * w[i]); 里面的应该是f[I-1][j]吧 少了个-1?