POJ 1011 - Java

這道題用的是depth first search，先對輸入排序，然後從長的stick開始配對結果。 剪枝非常關鍵，前3個剪枝缺一不可，否則都會超時。 其中第二個剪枝尤其容易被忽略：當能夠組成stick但是通過遞歸不能搜索到下一個stick怎麼組成的時候，這就表示已經沒辦法繼續了，

 

import java.util.Arrays; import java.util.Scanner; public class Main { static int[] fragments; static boolean[] used; static int total; static int max; static int num; public static void main(String[] args) { Scanner cin = new Scanner(System.in); String[] tokens; while ((num = Integer.valueOf(cin.nextLine())) != 0) { fragments = new int[num]; used = new boolean[num]; total = 0; max = 0; tokens = cin.nextLine().split(" "); for (int i = 0; i < num; i++) { fragments[i] = Integer.valueOf(tokens[i]); used[i] = false; total += fragments[i]; } Arrays.sort(fragments); max = fragments[num - 1]; for (; max <= total; max++) { if (total % max == 0) { int numSticks = total / max; if (dfs(0, num - 1, numSticks)) { System.out.println(max); } } } } } public static boolean dfs(int currentLength, int index, int numSticks) { if (currentLength == max) { numSticks--; currentLength = 0; index = num - 1; } if (numSticks == 0) { return true; } for (int i = index; i >= 0; i--) { if (used[i] == false) { if (currentLength + fragments[i] <= max) { used[i] = true; if (dfs(currentLength + fragments[i], i - 1, numSticks) == true) { return true; } used[i] = false; // remove branch #1: cannot even find solution to compose 1 stick if (currentLength == 0) { break; } // remove branch #2: this is the last solution you can find to compose a stick if (currentLength + fragments[i] == max) { break; } } else { // remove branch #3: skip all the adjacent fragments leading to same failure int k = i - 1; while (k >= 0 && fragments[k] == fragments[i]) { k--; i--; } } // remove branch #4: after skipping elements (remove branch #2), the total // length of all remaining fragments might be smaller than length required // // Note: the efficiency improvement gained by this action is not as huge as // the previous 3, so even if you omit this you can still get accepted int totalLeft = 0; int k = i - 1; while (k >=0) { totalLeft += fragments[k]; k--; } if ((totalLeft + currentLength) < max) { break; } } } return false; } }