(Greedy algorithm) Wooden Sticks

There is a pile of n wooden sticks. The length and weight of each stick is known in advance. The sticks is 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 is associated with cleaning operations and changing tools and shapes. The setup times of the woodworking machine is 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 would need no setup time for a stick of length l ' and weight W ' if l<=l ' and W<=w '. Otherwise, it'll need 1 minute for setup.

You is 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 is (4,9), (5,2), (2,1), (3,5), and (1,4), then the Minimum setup time should be 2 minutes since there are 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 on the first line of the input file. Each test case consists of a lines:the first line has a integer n, 1<=n<=5000, that represents the number of Wo Oden sticks in the test case, and the second line contains n 2 positive integers L1, W1, L2, W2, ..., LN, WN, each of Magn Itude at most 10000, where Li and wi is the length and weight of the i th wooden stick, respectively. The 2n integers is 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

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define MAXN 5010
using namespace Std;
int sum;
    struct node{int l;
int W;
}NODE[MAXN];
int VIS[MAXN];
    bool COM (const node &p,const node &q) {if (P.L&LT;=Q.L) return 1;
    if (P.L&LT;=Q.L&AMP;&AMP;P.W&LT;=Q.W) return 1;
    if (P.L==Q.L) return p.w<=q.w;

else return 0;
    } int main () {int cases;
    cin>>cases;
        while (cases--) {int n;
        int c=0;
        memset (vis,0,sizeof (VIS));
        sum=0;
        cin>>n;
            for (int i=0;i<n;++i) {cin>>node[i].l>>node[i].w;
        } sort (node,node+n,com);
        int flag=0;
        int BS;
                    for (int i=0;i<n;++i) {if (!vis[i]) {BS=NODE[I].W;
                    sum++;
                    Flag=i;
            Vis[i]=1; FoR (int j=i+1;j<n;++j) {if (Bs<=node[j].w&&!vis[j])
                            {vis[j]=1;
                        BS=NODE[J].W;
                }} I=flag;
    }} printf ("%d\n", sum); }
}