LeetCode 1029 Two City Scheduling (dp)

A company is planning to interview 2n people. Given the array costs where costs[i] = [aCosti, bCosti], the cost of flying the ith person to city a is aCosti, and the cost of flying the ith person to city b is bCosti.

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

Example 1:

Input: costs = [[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.

Example 2:

Input: costs = [[259,770],[448,54],[926,667],[184,139],[840,118],[577,469]]
Output: 1859

Example 3:

Input: costs = [[515,563],[451,713],[537,709],[343,819],[855,779],[457,60],[650,359],[631,42]]
Output: 3086

Constraints:

• 2 * n == costs.length
• 2 <= costs.length <= 100
• costs.length is even.
• 1 <= aCosti, bCosti <= 1000

Topic link ：https://leetcode.com/problems/two-city-scheduling/

The main idea of the topic ： Yes 2n personal , Everyone goes to the city A and B There are two cost, Let each city have its own n The minimum total number of people going cost

Topic analysis ： It's easy to think of a range of data n^3 Of dp,dp[i][a][b] To i personal , On the required page i The personal situation has gone down A City a personal , Went to the B City b Minimum total personal cost , Easy to get transfer equation ：

dp[i][a][b] = min(dp[i - 1][a - 1][b] + cost[i][0], dp[i - 1][a][b - 1] + cost[i][1])

because dp The equation determines each i Will choose at least one city , Therefore, one-dimensional space can be omitted

18ms, Time beats 5%

class Solution {
    public int twoCitySchedCost(int[][] costs) {
        int n = costs.length, m = n >> 1;
        int[][] dp = new int[n + 1][m + 1];
        for (int i = 1; i <= n; i++) {
            for (int a = 0; a <= m; a++) {
                for (int b = 0; b <= m; b++) {
                    if (a == 0 && b == 0) {
                        continue;
                    }
                    if (a + b <= i) {
                        if (a == 0) {
                            dp[i][a] = dp[i - 1][a] + costs[i - 1][1];
                        } else if (b == 0) {
                            dp[i][a] = dp[i - 1][a - 1] + costs[i - 1][0];
                        } else {
                            dp[i][a] = Math.min(dp[i - 1][a - 1] + costs[i - 1][0], dp[i - 1][a] + costs[i - 1][1]);
                        }
                        // System.out.println("dp["+i+"]["+a+"]["+b+"] = " + dp[i][a]);
                    }
                }
            }
        }
        return dp[n][m];
    }
}