LeetCode: Ones and Zeroes

题目：
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”.

题目给出m个0，n个1，问最多能组成多少个字符串。其中，组成的字符串应属于给出的字符串集合中的一个。
用动态规划来解。定义dp[i][j]为给出i个0，j个1时能组成的最大字符串数目，则动态转移函数为：
dp[i][j]=max{dp[i][j],dp[i-zeros][j-ones]+1}
其中zeros为字符串集合中某个字符串0的个数，相应地，ones为该字符串中1的个数。Accepted的代码：

class Solution {public:    int findMaxForm(vector<string>& strs, int m, int n) {        //dp[i][j]表示i个0,j个1时能得到的最大字符串数        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++)            {                if(strs[i][j]=='0') zeros++;                else if(strs[i][j]=='1') ones++;            }            for(int a=m;a>=zeros;a--)            {                for(int b=n;b>=ones;b--)                {                    if(dp[a][b]<dp[a-zeros][b-ones]+1)                        dp[a][b]=dp[a-zeros][b-ones]+1;                }            }        }        return dp[m][n];    }};