Leetcode 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.
以下是两个例子：
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".

其实这就是典型的二维费用背包问题，每个字符串都是由'0'或'1'构成的，而‘0’和‘1’的数量是有限的，每个字符串的价值又是相同的，那么问题就转化成每件物品价值都相同的价值的二维费用背包问题。而二维背包问题与一维背包问题的解决方法相似，只是维度不一样。

一维01背包问题的解决方法如下：

f[i][v] = Max{f[i-1][v],f[i-1][v-c[i]]+w[i]}      （*）

其中v为背包容量，f[i][v]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值，w[i]为第i件物品的价值，c[i]为第i件物品的成本。

其实这个过程就是一个从后往前推的过程，每一步考虑当前这件物品是否放到被包里，而（*）中第一项f[i-1][v]表示不取当前物品，后者则表示取当前物品。

本题代码如下：

class Solution {
private:
    int countNum(string str,char number){
        int ans = 0 ;
        for(string::size_type i = 0; i != str.size(); ++i){
            if(str[i] == number ) ++ans;
        }
        return ans;
    }
    vector<vector<int>> intiMat(vector<vector<int>> mat, int m, int n){
        for(int i = 0; i != m; ++i){
            vector<int> temp;
            for(int j = 0; j != n; ++j){
                temp.push_back(0);
            }
            mat.push_back(temp);
        }
		return mat;
    }
public:
    int findMaxForm(vector<string>& strs, int m, int n) {
        vector<vector<int>> mat;
        int n_zeros,n_ones;
        mat = intiMat(mat, m+1, n+1);
        for(vector<string>::iterator iter=strs.begin(); iter != strs.end(); ++iter){
            n_zeros = countNum(*iter,'0');
            n_ones = countNum(*iter,'1');
            for(int i = m; i >= n_zeros; --i){
                for(int j = n; j >= n_ones; --j){
                    mat[i][j] = (mat[i][j] > mat[i-n_zeros][j-n_ones]+1)?mat[i][j]:mat[i-n_zeros][j-n_ones]+1;
                }
            }
        }
        return mat[m][n];
    }
};