7-2 拼题A打卡奖励

#include<bits/stdc++.h>
using namespace std;

const int N = 1e3 + 10, M = 3e4 + 10;

int f[N][M]; //前i个物品获得j个金币奖励花费的最少时间

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int n, m;
    cin >> n >> m;
    vector<int> a(n + 1), b(n + 1);
    for(int i = 1; i <= n; i ++ ) cin >> a[i];
    int sum = 0;
    for(int i = 1; i <= n; i ++ ) cin >> b[i], sum += b[i];

    memset(f, 0x3f, sizeof f);
    for(int i = 0; i <= n; i ++ )  f[i][0] = 0;
    
    for(int i = 1; i <= n; i ++ )
    {
        for(int j = 1; j <= sum; j ++ )
        {
            f[i][j] = f[i - 1][j];
            if(j >= b[i])
            {
                f[i][j] = min(f[i][j], f[i - 1][j - b[i]] + a[i]);
            }
        }
    }
    
    int ans = 0;
    for(int i = 1; i <= sum; i ++ )
    {
        if(f[n][i] <= m) ans = i;
    }
    cout << ans << endl;
    return 0;
}

空间优化版：

#include<bits/stdc++.h>
using namespace std;

const int N = 1e3 + 10, M = 3e4 + 10;

int f[M]; //前i个物品获得j个金币奖励花费的最少时间

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int n, m;
    cin >> n >> m;
    vector<int> a(n + 1), b(n + 1);
    for(int i = 1; i <= n; i ++ ) cin >> a[i];
    int sum = 0;
    for(int i = 1; i <= n; i ++ ) cin >> b[i], sum += b[i];

    memset(f, 0x3f, sizeof f);
    f[0] = 0;
    
    for(int i = 1; i <= n; i ++ )
        for(int j = sum; j >= b[i]; j -- )
                f[j] = min(f[j], f[j - b[i]] + a[i]);
    
    int ans = 0;
    for(int i = 1; i <= sum; i ++ ) if(f[i] <= m) ans = i;
    cout << ans << endl;
    return 0;
}