[UVa] Wooden Sticks

Question

link

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).

Input

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, ..., l_n, w_n, each of magnitude at most 10000, where l_i and w_i are the length and weight of the i th wooden stick, respectively. The 2n integers are delimited by one or more spaces.

Output 

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

Sample Input

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

Sample Output 

2
1
3

Analysis

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

Solution

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

Code

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) {
            p++;
        }
        if (p == len) {
            break;
        }
        // use points[p] as the first elements in the queue
        Pair temp = copy(points[p]);
        points[p] = null;
        total++;

        // 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) {
                continue;
            }
            temp = copy(points[k]);
            points[k] = null;
        }
        p++;
    }
    return total;
}

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

class PointComparator implements Comparator<Pair> {

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