ACM-problem I

problem：

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

不说啥了，和problem B一样

代码：

#include <iostream>
#include<algorithm>
using namespace std;
struct stick
{int longs;
int weight;
}ST[5001];
int f[5001];
int cmp(const stick &a,const stick &b)
{if(a.longs!=b.longs)
return a.longs>b.longs;
else
return a.weight>b.weight;
}
int main()
{int T,n;
int t,i,j;
int tim;


cin>>T;
while(T--)
{
cin>>n;
for( i=0;i<n;i++)
{cin>>ST[i].longs>>ST[i].weight;
 f[i]=0;
}
sort(ST,ST+n,cmp);
 tim=0;
for ( i=0;i<n;i++)
{if(f[i]) continue;
t=ST[i].weight;
for(int j=i+1;j<n;j++)
{
  if(t>=ST[j].weight&&!f[j])
 {
  t=ST[j].weight;
  f[j]=1;
  }
 }
tim++;
}
cout<<tim<<endl;
}


return 0;
}