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 stickif (currentLength == 0) {break;}// remove branch #2: this is the last solution you can find to compose a stickif (currentLength + fragments[i] == max) {break;}} else {// remove branch #3: skip all the adjacent fragments leading to same failureint 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 acceptedint totalLeft = 0;int k = i - 1;while (k >=0) {totalLeft += fragments[k];k--;}if ((totalLeft + currentLength) < max) {break;}}}return false;}}