(贪心算法)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

#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<=q.l)
        return 1;
    if(p.l<=q.l&&p.w<=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);
    }
}