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”.

动态规划：dp[i][j]表示i个0，j个1最多能组成多少个给定数组中的字符串，假如数组中一个字符串为str，str中0的个数为zeros，1的个数为ones，则状态转移方程为dp[i][j] = max(dp[i-zeros][j-ones] + 1,dp[i][j])，从后往前更新dp数组。

class Solution {
public:
    int findMaxForm(vector<string>& strs, int m, int n) {
        vector<vector<int>> dp(m + 1,vector<int>(n + 1,0));
    for(int i = 0;i < strs.size(); i++){
        int zeros = 0,ones = 0;
        for(int j = 0;j < strs[i].length();j++)
            strs[i][j] == '0' ? zeros ++ : ones ++;
        for(int j = m;j >= zeros;j--)
            for(int k = n;k >= ones;k--)
                dp[j][k] = max(dp[j - zeros][k-ones] + 1,dp[j][k]);
    }
    return dp[m][n];
    }
};