Woodstock Blog

a tech blog for general algorithmic interview questions

[UVa] Wooden Sticks



There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to prepare processing a stick. The setup times are associated with cleaning operations and changing tools and shapes in the machine. The setup times of the woodworking machine are given as follows:

  • The setup time for the first wooden stick is 1 minute.
  • Right after processing a stick of length l and weight w , the machine will need no setup time for a stick of length l' and weight w' if l <=l' and w <=w' .

Otherwise, it will need 1 minute for setup.
You are to find the minimum setup time to process a given pile of n wooden sticks. For example, if you have five sticks whose pairs of length and weight are (4,9), (5,2), (2,1), (3,5), and (1,4) then the minimum setup time should be 2 minutes since there is a sequence of pairs (1,4), (3,5), (4,9), (2,1), (5,2).


The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case consists of two lines: The first line has an integer n, 1<=n<=5000, that represents the number of wooden sticks in the test case, and the second line contains 2n positive integers l1, w1, l2, w2, …, ln, wn, each of magnitude at most 10000, where li and wi are the length and weight of the i th wooden stick, respectively. The 2n integers are delimited by one or more spaces.


The output should contain the minimum setup time in minutes, one per line.

Sample Input

4 9 5 2 2 1 3 5 1 4
2 2 1 1 2 2
1 3 2 2 3 1

Sample Output 



This question is a little similar to [CC150v5] 9.10 Stack Up the Boxes, yet different.


Solution quoted from 脚本百事通:

按长度从小到大排序, 若长度相同, 则按重量从小到大排序(先按重量再按长度也行)

然后, 我们针对当前木棍, 与剩下的木棍比较, 满足 w1 <= w2 && l1 <= l2 的, 就更新一下当前棍, 并标记~

当所有木棍都被编组后 输出到底设置了几次~

ref1, ref2


The following Java code is written by me

public int count(Pair[] points) {
    Arrays.sort(points, new PointComparator());
    // now the array is sorted with Point.x
    int total = 0;
    int p = 0;
    int len = points.length;

    while (p < len) {
        // find the next non-null point
        while (p < len && points[p] == null) {
        if (p == len) {
        // use points[p] as the first elements in the queue
        Pair temp = copy(points[p]);
        points[p] = null;

        // then mark all elements that follow points[p]
        for (int k = p + 1; k < len; k++) {
            if (points[k] == null || points[k].y < temp.y) {
            temp = copy(points[k]);
            points[k] = null;
    return total;

Pair copy(Pair p) {
    Pair newP = new Pair(p.x, p.y);
    return newP;

class PointComparator implements Comparator<Pair> {

    public int compare(Pair p1, Pair p2) {
        return p1.x - p2.x;