Leetcode 1029 Two City Scheduling (两个城市调度) (Greey, DP)

Leetcode 1029

题目描述

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.

例子

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.

解题思路

  方法一：(贪心算法)
  按照cost[0]-cost[1]排序，前N个fly to A，后N个fly to B

Python
class Solution:
    def twoCitySchedCost(self, costs):
        N = len(costs)
        dummy = []
        for cost in costs:
            dummy.append([cost, cost[0]-cost[1]])
        dummy = sorted(dummy, key=(lambda x : x[1]))
        ans = 0
        for i in range(N):
            ans += dummy[i][0][0] if i< N>>1 else dummy[i][0][1]
        return ans

Java
class Solution {
    public int twoCitySchedCost(int[][] costs){
        Arrays.sort(costs, new Comparator<int[]>(){
            public int compare(int[] a, int[] b){
                return (a[0]+b[1]) - (a[1]+b[0]);
            }
        });
        int ans = 0;
        for(int i=0; i<costs.length; ++i){
            ans += i < costs.length>>1? costs[i][0]:costs[i][1];
        }
        return ans;
    }
    }
}

  方法二：(动态规划)

Python
class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        N = len(costs)>>1
        dp = [[0 for _ in range(N+1)] for _ in range(N+1)]
        for i in range(1,N+1):
            dp[i][0] = dp[i-1][0]+costs[i-1][0]
        for j in range(1,N+1):
            dp[0][j] = dp[0][j-1]+costs[j-1][1]
        for i in range(1,N+1):
            for j in range(1,N+1):
                dp[i][j] = min(dp[i-1][j]+costs[i+j-1][0], dp[i][j-1]+costs[i+j-1][1])
        return dp[-1][-1]

Java
class Solution {
    public int twoCitySchedCost(int[][] costs) {
        int N = costs.length >> 1;
        int[][] dp = new int[N + 1][N + 1];
        for (int i = 1; i <= N; ++i) {
            dp[i][0] = dp[i - 1][0] + costs[i - 1][0];
        }
        for (int j = 1; j <= N; ++j) {
            dp[0][j] = dp[0][j - 1] + costs[j - 1][1];
        }
        for (int i = 1; i <= N; ++i) {
            for (int j = 1; j <= N; ++j) {
                dp[i][j] = Math.min(dp[i - 1][j] + costs[i + j - 1][0], dp[i][j - 1] + costs[i + j - 1][1]);
            }
        }
        return dp[N][N];
    }
}