1051 Wooden Sticks

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Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4694 Accepted Submission(s): 1933

Problem Description

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<iostream>
#include<algorithm>
using namespace std;
typedef struct Node
{
 int len,w;
}Node;
Node p[5010];
int f[5010];
bool cmp(Node a,Node b)
{
 if(a.len!=b.len) return a.len<b.len;
 return a.w<b.w;
}
int main()
{
 int t,i,j,n;
 cin>>t;
 while(t--)
 {
  cin>>n;
  for(i=0;i<n;i++)
   cin>>p[i].len>>p[i].w;
  sort(p,p+n,cmp);
  int cnt=0;
  for(i=0;i<n;i++) f[i]=1;
  for(i=1;i<n;i++)
   for(j=0;j<i;j++)
    if(p[j].w>p[i].w&&f[i]<f[j]+1) f[i]=f[j]+1;
    int sum=0;
    for(i=0;i<n;i++) if(f[i]>sum) sum=f[i];
    cout<<sum<<endl;
 }
 return 0;
}