LeetCode刷题之——1029. Two City Scheduling (双城面试2N人费用最小问题)

1029. Two City Scheduling (双城面试2N人费用最小问题)

There are 2N people a company is planning to interview. The cost of flying the i-th person to city A is costs[i][0], and the cost of flying the i-th person to city B is costs[i][1].

Return the minimum cost to fly every person to a city such that exactly N people arrive in each city.

Example 1:

Input: [[10,20],[30,200],[400,50],[30,20]]
Output: 110
Explanation: 
The first person goes to city A for a cost of 10.
The second person goes to city A for a cost of 30.
The third person goes to city B for a cost of 50.
The fourth person goes to city B for a cost of 20.

The total minimum cost is 10 + 30 + 50 + 20 = 110 to have half the people interviewing in each city.

Note:

1. 1 <= costs.length <= 100
2. It is guaranteed that costs.length is even.
3. 1 <= costs[i][0], costs[i][1] <= 1000

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以上是原题描述。

现在来说一下我的算法：

1. 从每个数对中选择较小的那个，选完后分成了a、b两列
2. 看a和b谁更多，计算出多的一方需要补偿给对方几个（n）
3. 计算多的一列中的数字在原来数对中的差值，选择前n个差值最小的数对，进行替换

比如上面的例子，第一步后，a列：259；b列：54，667，139，118，469.

第二步：a列1个，b列5个，相差4个，b需要补给a列2个。

第三步：计算出新的差值列：394，259，45，722，108. 发现最小的2个是45和108，对应的是184替换139，577替换469.（见下图）

 最终我们选择的数字是：259，54，667，184，139，118，577.

和为1859.

代码：

class Solution {
public:
    int twoCitySchedCost(vector<vector<int>>& costs) {
        int result=0;
        vector<vector<int>> a;//左列
        vector<vector<int>> b;//右列
        vector<int> c;          //辅助
        vector<vector<int>> d;//差值
        int diff;
        
        //初始分选，每个数对选最小的那个
        for (auto iter = costs.begin(); iter != costs.end(); iter++)
        {
         if((*iter)[0]>(*iter)[1]) {
             c.push_back((*iter)[1]);
             c.push_back(iter-costs.begin());//返回原始数组的第一维下标
             b.push_back(c);
         }
            else{
                c.push_back((*iter)[0]);
                c.push_back(iter-costs.begin());//返回原始数组的第一维下标
                a.push_back(c);
            }
            c.clear();
        }
        
        diff=a.size()-b.size();//a比b多几个
        if(diff==0){
              for (auto iter = a.begin(); iter != a.end(); iter++)
                result+=(*iter)[0] ;
     
              for (auto iter = b.begin(); iter != b.end(); iter++)
                result+=(*iter)[0] ;
        }
        else if(diff > 0){//a>b
            diff = diff/2;//2N人一定能整除
            for (auto iter = a.begin(); iter != a.end(); iter++)
            {
                int index=(*iter)[1];
                c.push_back(abs(costs[index][0]-costs[index][1]));
                c.push_back(index);
                d.push_back(c);
                c.clear();
            } 
            for (auto iter = b.begin(); iter != b.end(); iter++)
                result+=(*iter)[0] ;
            while(diff-->0){
                int index = smallest(d);//注意这里
                 for (auto iter = a.begin(); iter != a.end(); iter++)
                {
                    if((*iter)[1]==index)
                        (*iter)[0]=costs[(*iter)[1]][1];//注意第二个下标是1
                }
            }
            for (auto iter = a.begin(); iter != a.end(); iter++)
                result+=(*iter)[0] ;
        }
        else{//diff<0
            //a<b
            diff=0-diff;
            diff = diff/2;
            for (auto iter = b.begin(); iter != b.end(); iter++)
            {
                int index=(*iter)[1];
                c.push_back(abs(costs[index][0]-costs[index][1]));
                c.push_back(index);//返回原始数组的第一维下标
                d.push_back(c);
                c.clear();
            } 
            for (auto iter = a.begin(); iter != a.end(); iter++)
                result+=(*iter)[0] ;
            while(diff>0){
                int index = smallest(d);//注意这里
                 for (auto iter = b.begin(); iter != b.end(); iter++)
            {
                    if((*iter)[1]==index){
                        (*iter)[0]=costs[(*iter)[1]][0];//注意第二个下标是0
                        diff--;
                    }
                        
                    
            }
            }
           
            for (auto iter = b.begin(); iter !=b.end(); iter++)
                result+=(*iter)[0] ;
        }
        
        return result;
    }
    
    int  smallest(vector<vector<int>>& hahaha){
        vector<int> c;
        for (auto iter = hahaha.begin(); iter != hahaha.end(); iter++)
            c.push_back((*iter)[0]);
        auto smallest=min_element(begin(c),end(c));
        int index = hahaha[distance(begin(c),smallest)][1];
        hahaha[distance(begin(c),smallest)][0]=1000;
        return index;
    }
};

本地测试用的主函数：

int main() {
    Solution* mySo = new Solution();
    vector<vector<int>> cos = { {259,770},{448,54},{926,667},{184,139},{840,118},{577,469} };
    cout << mySo->twoCitySchedCost(cos) << endl;
}

注：这道题难度是easy·········