leetcode 474. Ones and Zeroesのいくつかの0と1の構成文字列の最大数+動的計画DP+リュックサックの問題

In the computer world, use restricted resource you have to generate maximum benefit is what we always want to pursue.
For now, suppose you are a dominator of m 0s and n 1s respectively. On the other hand, there is an array with strings consisting of only 0s and 1s.
Now your task is to find the maximum number of strings that you can form with given m 0s and n 1s. Each 0 and 1 can be used at most once.
Note: The given numbers of 0s and 1s will both not exceed 100 The size of given string array won’t exceed 600. Example 1: Input: Array = {“10”, “0001”, “111001”, “1”, “0”}, m = 5, n = 3 Output: 4
Explanation: This are totally 4 strings can be formed by the using of 5 0s and 3 1s, which are “10,”0001”,”1”,”0” Example 2: Input: Array = {“10”, “0”, “1”}, m = 1, n = 1 Output: 2
Explanation: You could form “10”, but then you’d have nothing left. Better form “0” and “1”.
この問題は実はとても簡単で、どのように最初の時どのようにすることを思い出せないで、本題は1つのリュックサックの問題の改版で、見たところ学習と進歩の地方が本当に多すぎて、格差はまだ大きいです
そしてleetcode 494.Target Sumターゲットと+リュックサック問題+深さ優先DFS+ダイナミックプランニングDP+簡単な解析的導出
また、leetcode 518も提案する.Coin Change 2動的計画DP、leetcode 279.Perfect Squaresリュックサックのような問題+簡単なダイナミックプランニングDP解決、leetcode 377.Combination Sum IVの組合せの和+DPダイナミックプランニング+DFS深さ優先ループとleetcode 416.Partition Equal Subset SumダイナミックプランニングDP+深さ優先DFSを巡る学習
コードは次のとおりです.

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

class Solution 
{
public:
    int findMaxForm(vector<string>& strs, int m, int n) 
    {
        //dp[i][j]   i 0 j 1           
        vector<vector<int>> dp(m+1,vector<int>(n+1,0));
        for (string s : strs)
        {
            int numOne = 0, numZero = 0;
            for (char c : s)
            {
                if (c == '0')
                    numZero++;
                else if (c == '1')
                    numOne++;
            }

            for (int i = m; i >= numZero; i--)
            {
                for (int j = n; j >= numOne; j--)
                {
                    dp[i][j] = max(dp[i][j], dp[i - numZero][j - numOne] + 1);
                }
            }

        }
        return dp[m][n];
    }
};