1022. 宠物小精灵之收服

思路

双层dp

代码

#include <bits/stdc++.h>
using namespace std;const int N = 1010, mod = 1e9 + 7;int n, m, k, x, y, z, ans, t;
int w[N], f[N][N];void solve()
{cin >> n >> m >> k;for (int i = 1; i <= k; i ++ ){cin >> x >> y;for (int j = n; j >= x; j -- ){for (int l = m - 1; l >= y; l -- ){f[j][l] = max(f[j][l], f[j - x][l - y] + 1);}}}cout << f[n][m - 1] << " ";ans = m - 1;while (ans > 0 && f[n][ans - 1] == f[n][m - 1]) ans --;cout << m - ans << "\n";
}signed main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T = 1;
//	cin >> T;while (T -- ){solve();}
}