474. 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”.

解答：

转移方程：

这其实是个二维背包问题。令dp[j][k]表示还剩j个0，k个1时最大可以放多少个01字符串。z[i]表示第i个字符串中0的个数，o[i]表示第i个字符串中1的个数。关键在于这第i个01字符串是放还是不放。

dp[j][k] = max{dp[j][k], dp[j - z[i]][k - o[i]] + 1}

bitset的使用——本文用以统计字符串中出现的1的个数

代码：

class Solution {public:    int findMaxForm(vector<string>& strs, int m, int n) {        int num = strs.size();  // 共多少个字符串        // 统计出所有01字符串中的0、1个数        vector<int> zeros;        vector<int> ones;        for (auto i : strs) {            bitset<200> tmp(i);            ones.push_back(tmp.count());            zeros.push_back(i.size() - tmp.count());        }        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));  // 0 个‘0' 到 m 个'0'        for (int i = 0; i < num; i++) {            for (int j = m; j >= zeros[i]; j--) {                for (int k = n; k >= ones[i]; k--) {                    dp[j][k] = max(dp[j][k], dp[j - zeros[i]][k - ones[i]] + 1);                }            }        }        return dp[m][n];    }};int main() {    vector<string> strs = { "10", "0001", "111001", "1", "0" };    int m = 5;    int n = 3;    Solution solution;    int res = solution.findMaxForm(strs, m, n);    cout << res << endl;    system("pause");    return 0;}