56 lines
1.9 KiB
C++
56 lines
1.9 KiB
C++
// Source : https://leetcode.com/problems/two-city-scheduling/
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// Author : Hao Chen
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// Date : 2019-04-21
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/*****************************************************************************************************
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*
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* There are 2N people a company is planning to interview. The cost of flying the i-th person to city
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* A is costs[i][0], and the cost of flying the i-th person to city B is costs[i][1].
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*
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* Return the minimum cost to fly every person to a city such that exactly N people arrive in each
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* city.
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*
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* Example 1:
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*
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* Input: [[10,20],[30,200],[400,50],[30,20]]
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* Output: 110
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* Explanation:
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* The first person goes to city A for a cost of 10.
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* The second person goes to city A for a cost of 30.
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* The third person goes to city B for a cost of 50.
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* The fourth person goes to city B for a cost of 20.
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*
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* The total minimum cost is 10 + 30 + 50 + 20 = 110 to have half the people interviewing in each city.
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*
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* Note:
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*
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* 1 <= costs.length <= 100
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* It is guaranteed that costs.length is even.
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* 1 <= costs[i][0], costs[i][1] <= 1000
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******************************************************************************************************/
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class Solution {
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private:
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static int diff(vector<int>& x) {
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return x[1] - x[0];
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}
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static bool cmpfunc(vector<int>& lhs, vector<int>& rhs) {
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return diff(lhs) > diff(rhs);
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}
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public:
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// Just simply sort the array by comparing the different cost go to A city and B city
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// then the bigger difference would be in left and right side, and the smaller difference would be in the middle
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// We could simply let the first half go to A city, and the second half go to B city.
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int twoCitySchedCost(vector<vector<int>>& costs) {
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sort(costs.begin(), costs.end(), cmpfunc);
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int result = 0;
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int len = costs.size();
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for (int i=0; i<len/2; i++) {
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result += (costs[i][0] + costs[len-i-1][1]);
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}
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return result;
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}
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};
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