JOJ1048: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: 


(a) The setup time for the first wooden stick is 1 minute. 
(b) 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 n 2 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.
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

思路：

首先随便选一个元素当第一关键字，另一个为第二关键字进行快排。

之后我们只看第二关键字，这道题就变成了拦截导弹的第二问，最少的装置数。

根据偏序集的Dilworth定理，偏序集的最大反链的长度等于链的划分的最小数目。

所以，我们要求的非降序的最小数目，也就是它的最大反链的长度。

代码：

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;struct node{    int w, l;} t[10002];int cmp(const void *a, const void *b){    node *x = (node *)a;    node *y = (node *)b;    if (x -> l == y -> l)        return x -> w - y -> w;    else        return x -> l - y -> l;}int main(){    int cn;    cin >> cn;    for (int ci = 1; ci <= cn; ++ci)    {        int n, f[10001] = {0}, max = 1;        cin >> n;        for (int i = 0; i < n; ++i)            cin >> t[i].l >> t[i].w;        qsort(t, n, sizeof(t[0]), cmp);        f[0] = 1;        for (int i = 1; i < n; ++i)        {            f[i] = 1;            for (int j = 0; j < i; ++j)                if (f[j] + 1 > f[i] && t[j].w > t[i].w)                {                    f[i] = f[j] + 1;                    if (f[i] > max)                        max = f[i];                }        }        cout << max << endl;    }    return 0;}