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| 1 | +package dynamic_programming; |
| 2 | + |
| 3 | +/** |
| 4 | + * Created by gouthamvidyapradhan on 13/04/2019 |
| 5 | + * |
| 6 | + * LeetCode wants to give one of its best employees the option to travel among N cities to collect algorithm |
| 7 | + * problems. But all work and no play makes Jack a dull boy, you could take vacations in some particular cities and |
| 8 | + * weeks. Your job is to schedule the traveling to maximize the number of vacation days you could take, but there are |
| 9 | + * certain rules and restrictions you need to follow. |
| 10 | + * |
| 11 | + * Rules and restrictions: |
| 12 | + * You can only travel among N cities, represented by indexes from 0 to N-1. Initially, you are in the city indexed 0 |
| 13 | + * on Monday. |
| 14 | + * The cities are connected by flights. The flights are represented as a N*N matrix (not necessary symmetrical), |
| 15 | + * called flights representing the airline status from the city i to the city j. If there is no flight from the city |
| 16 | + * i to the city j, flights[i][j] = 0; Otherwise, flights[i][j] = 1. Also, flights[i][i] = 0 for all i. |
| 17 | + * You totally have K weeks (each week has 7 days) to travel. You can only take flights at most once per day and can |
| 18 | + * only take flights on each week's Monday morning. Since flight time is so short, we don't consider the impact of |
| 19 | + * flight time. |
| 20 | + * For each city, you can only have restricted vacation days in different weeks, given an N*K matrix called days |
| 21 | + * representing this relationship. For the value of days[i][j], it represents the maximum days you could take |
| 22 | + * vacation in the city i in the week j. |
| 23 | + * You're given the flights matrix and days matrix, and you need to output the maximum vacation days you could take |
| 24 | + * during K weeks. |
| 25 | + * |
| 26 | + * Example 1: |
| 27 | + * Input:flights = [[0,1,1],[1,0,1],[1,1,0]], days = [[1,3,1],[6,0,3],[3,3,3]] |
| 28 | + * Output: 12 |
| 29 | + * Explanation: |
| 30 | + * Ans = 6 + 3 + 3 = 12. |
| 31 | + * |
| 32 | + * One of the best strategies is: |
| 33 | + * 1st week : fly from city 0 to city 1 on Monday, and play 6 days and work 1 day. |
| 34 | + * (Although you start at city 0, we could also fly to and start at other cities since it is Monday.) |
| 35 | + * 2nd week : fly from city 1 to city 2 on Monday, and play 3 days and work 4 days. |
| 36 | + * 3rd week : stay at city 2, and play 3 days and work 4 days. |
| 37 | + * Example 2: |
| 38 | + * Input:flights = [[0,0,0],[0,0,0],[0,0,0]], days = [[1,1,1],[7,7,7],[7,7,7]] |
| 39 | + * Output: 3 |
| 40 | + * Explanation: |
| 41 | + * Ans = 1 + 1 + 1 = 3. |
| 42 | + * |
| 43 | + * Since there is no flights enable you to move to another city, you have to stay at city 0 for the whole 3 weeks. |
| 44 | + * For each week, you only have one day to play and six days to work. |
| 45 | + * So the maximum number of vacation days is 3. |
| 46 | + * Example 3: |
| 47 | + * Input:flights = [[0,1,1],[1,0,1],[1,1,0]], days = [[7,0,0],[0,7,0],[0,0,7]] |
| 48 | + * Output: 21 |
| 49 | + * Explanation: |
| 50 | + * Ans = 7 + 7 + 7 = 21 |
| 51 | + * |
| 52 | + * One of the best strategies is: |
| 53 | + * 1st week : stay at city 0, and play 7 days. |
| 54 | + * 2nd week : fly from city 0 to city 1 on Monday, and play 7 days. |
| 55 | + * 3rd week : fly from city 1 to city 2 on Monday, and play 7 days. |
| 56 | + * Note: |
| 57 | + * N and K are positive integers, which are in the range of [1, 100]. |
| 58 | + * In the matrix flights, all the values are integers in the range of [0, 1]. |
| 59 | + * In the matrix days, all the values are integers in the range [0, 7]. |
| 60 | + * You could stay at a city beyond the number of vacation days, but you should work on the extra days, which won't be |
| 61 | + * counted as vacation days. |
| 62 | + * If you fly from the city A to the city B and take the vacation on that day, the deduction towards vacation days |
| 63 | + * will count towards the vacation days of city B in that week. |
| 64 | + * We don't consider the impact of flight hours towards the calculation of vacation days. |
| 65 | + * |
| 66 | + * Solution: |
| 67 | + * O(N x N x K) |
| 68 | + * Start from the last week K; Calculate for the last week maximum vacation days for all possible cities. |
| 69 | + * Now, iteratively calculate backwards |
| 70 | + * for each week K - 1 |
| 71 | + * and for each city |
| 72 | + * and for each available connection from the city |
| 73 | + * memoize the maximum vacation days possible for this city |
| 74 | + * Return the answer for city 0 and week 0 |
| 75 | + */ |
| 76 | +public class MaximumVacationDays { |
| 77 | + |
| 78 | + /** |
| 79 | + * Main method |
| 80 | + * @param args |
| 81 | + */ |
| 82 | + public static void main(String[] args) { |
| 83 | + int[][] flights = {{0,1,0,1,1},{1,0,0,1,1},{1,0,0,1,0},{0,1,0,0,0},{0,0,0,1,0}}; |
| 84 | + int[][] days = {{0,4,1,6,6},{4,3,3,0,1},{3,6,6,5,0},{1,3,1,1,4},{5,3,3,3,4}}; |
| 85 | + System.out.println(new MaximumVacationDays().maxVacationDays(flights, days)); |
| 86 | + } |
| 87 | + |
| 88 | + public int maxVacationDays(int[][] flights, int[][] days) { |
| 89 | + int N = days.length; |
| 90 | + int W = days[0].length; |
| 91 | + int[][] DP = new int[N][W + 1]; |
| 92 | + for(int w = W - 1; w >= 0; w --){ |
| 93 | + for(int n = 0; n < N; n ++){ |
| 94 | + DP[n][w] = days[n][w] + DP[n][w + 1]; |
| 95 | + } |
| 96 | + |
| 97 | + for(int n = 0; n < N; n ++){ |
| 98 | + int max = Integer.MIN_VALUE; |
| 99 | + int[] F = flights[n]; |
| 100 | + for(int i = 0; i < F.length; i ++){ |
| 101 | + if(F[i] == 1){ |
| 102 | + max = Math.max(max, days[i][w] + DP[i][w + 1]); |
| 103 | + } |
| 104 | + } |
| 105 | + DP[n][w] = Math.max(DP[n][w], max); |
| 106 | + } |
| 107 | + } |
| 108 | + return DP[0][0]; |
| 109 | + } |
| 110 | +} |
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