package experience13; import java.util.Arrays; public class Load { // 集装箱数 static int n; // 集装箱重量数组 static int[] weightArray; // 第一艘轮船的载重量 static int c; // 当前载重量 static int currentWeight; // 当前最优载重量 static int bestWeight; // 剩余集装箱重量 static int rest; // 当前解 static int[] currentSolution; // 当前最优解 static int[] bestCurrentSolution; public static int maxLoading(int[] ww, int cc, int[] xx) { n = ww.length; weightArray = ww; c = cc; currentWeight = 0; bestWeight = 0; currentSolution = new int[n]; bestCurrentSolution = xx; for (int i = 0; i < n; i++) { rest += weightArray[i]; } backtrack(0); return bestWeight; } public static void backtrack(int i) { if (i >= n) { if (currentWeight > bestWeight) { for (int j = 0; j < n; j++) { bestCurrentSolution[j] = currentSolution[j]; } bestWeight = currentWeight; } return; } rest -= weightArray[i]; if (currentWeight + weightArray[i] <= c) { currentSolution[i] = 1; currentWeight += weightArray[i]; backtrack(i + 1); currentWeight -= weightArray[i]; } if (currentWeight + rest > bestWeight) { currentSolution[i] = 0; backtrack(i + 1); } rest += weightArray[i]; } public static void main(String[] args) { weightArray = new int[]{5, 2, 1, 3}; c = 10; int[] array = new int[4]; System.out.println(maxLoading(weightArray, c, array)); System.out.println(Arrays.toString(array)); } }分析算法时间复杂性

该算法使用了回溯法求解装载问题，时间复杂度取决于搜索树的大小。具体来说，搜索树的节点数为 $2^n$，其中 $n$ 为集装箱数。因此，该算法的时间复杂度为 $O(2^n)$。

在实际应用中，如果集装箱数较多，该算法的时间复杂度会非常高，因此需要采用更加高效的算法来解决装载问题。例如，可以使用动态规划、贪心算法等方法来降低时间复杂度。