http://acm.hdu.edu.cn/showproblem.php?pid=1051
Wooden Sticks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10738 Accepted Submission(s): 4422
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
#include<iostream>
#include<stdlib.h>
#include<string.h>
int c[5200];
using namespace std;
struct node
{
int l,w;
}s[5200];
int cmp( const void *a , const void *b )
{
struct node *c = (struct node *)a;
struct node *d = (struct node *)b;
if( c->l != d->l ) return c->l - d->l ;
else return c->w - d->w ;
}
int main()
{
int cishu,k;
cin>>cishu;
for(k=1;k<=cishu;k++)
{
int n,i,j,sum,weight;
sum=0;
cin>>n;
for(i=0;i<n;i++)
{
cin>>s[i].l>>s[i].w;
c[i]=0;
}
qsort(s,n,sizeof(s[0]),cmp);
for(i=0;i<n;i++)
{
if(c[i]==0)
{
sum=sum+1;
c[i]=1;
weight=s[i].w;
for(j=i+1;j<n;j++)
{
if(c[j]==0&&s[j].w>=weight)
{
c[j]=1;
weight=s[j].w;
}
}
}
}
cout<<sum<<endl;
}
return 0;
}