474. Ones and Zeroes

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:
1. The given numbers of 0s and 1s will both not exceed 100
2. The size of given string array won’t exceed 600.

Example 1:Input: Array = {"10", "0001", "111001", "1", "0"}, m = 5, n = 3Output: 4Explanation: 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 = 1Output: 2Explanation: You could form "10", but then you'd have nothing left. Better form "0" and "1".

此题用回溯法做了一下，做的时候就怀着TLR的心情，果然如此。看出了是动态规划的题目，但是状态转移方程写不出来啊，老铁。。。
贴出参考了别人代码之后的代码吧；
这道题是一道典型的应用DP来解的题，如果我们看到这种求总数，而不是列出所有情况的题，十有八九都是用DP来解，重中之重就是在于找出递推式。如果你第一反应没有想到用DP来做，想得是用贪心算法来做，比如先给字符串数组排个序，让长度小的字符串在前面，然后遍历每个字符串，遇到0或者1就将对应的m和n的值减小，这种方法在有的时候是不对的，比如对于{“11”, “01”, “10”}，m=2，n=2这个例子，我们将遍历完“11”的时候，把1用完了，那么对于后面两个字符串就没法处理了，而其实正确的答案是应该组成后面两个字符串才对。
所以我们需要建立一个二位的DP数组，其中dp[i][j]表示有i个0和j个1时能组成的最多字符串的个数，而对于当前遍历到的字符串，我们统计出其中0和1的个数为zeros和ones，然后dp[i - zeros][j - ones]表示当前的i和j减去zeros和ones之前能拼成字符串的个数，那么加上当前的zeros和ones就是当前dp[i][j]可以达到的个数，我们跟其原有数值对比取较大值即可，所以递推式如下：

dp[i][j] = max(dp[i][j], dp[i - zeros][j - ones] + 1);

//回溯法带来的时间复杂度急剧上升，行不通！！！int dfs_MaxForm1(vector<string>& strs, int pos, int m, int n){    if (m <= 0 && n <= 0)return 0;    if (pos == strs.size())return 0;    int m0 = 0, n1 = 0, cnt1 = 0, cnt2 = 0;    count(strs[pos], m0, n1);    if (m - m0 >= 0 && n - n1 >= 0){        cnt1 = dfs_MaxForm1(strs, pos + 1, m - m0, n - n1) + 1;    }    cnt2 = dfs_MaxForm1(strs, pos + 1, m, n);    return max(cnt2, cnt1);}//DPint findMaxForm(vector<string>& strs, int m, int n) {    if (strs.size() == 0 || (m == 0 && n == 0))return 0;    vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));    for (auto x : strs){        int zeros = 0, ones = 0;        for (auto y : x)(y == '0') ? zeros++ : ones++;        for (int i = m; i >= zeros; i--){            for (int j = n; j >= ones; j--){                dp[i][j] = max(dp[i][j], dp[i - zeros][j - ones] + 1);            }        }    }    return dp[m][n];}