Wooden Sticks

E - Wooden Sticks

Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

SubmitStatus

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

 213 

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int main(){    int n,m,l[5005],w[5005], t[5005],u[5005],d; int e[5005],r=0;    scanf("%d",&m);    while(m--)    {    memset(t,0,sizeof(t));        memset(u,0,sizeof(u));        memset(l,0,sizeof(l));        memset(w,0,sizeof(w));    scanf("%d",&n);    for(int b=0;b<n;b++)       scanf("%d%d",&l[b],&w[b]);//cin>>l[b]>>w[b];     for(int j=0;j<n-1;j++)       for(int k=j+1;k<n;k++)         {        if(l[j]>l[k])          {          d=l[j];l[j]=l[k];l[k]=d;           d=w[j];w[j]=w[k];w[k]=d;          }          else if(l[j]==l[k])            {               if(w[j]>w[k])   {   d=w[j];w[j]=w[k];w[k]=d;      }             }        }         int count,x=0 ;       for(int i=0;i<n;i++)           {                                              for(int j=0;j<=x;j++){    count =0;  if(t[j]<=l[i]&&u[j]<=w[i])    {t[j]=l[i];u[j]=w[i];count=1;break;}        }    if(count ==0)   {                x++;t[x]=l[i];u[x]=w[i];                      }                        }if(n==0) x=-1; e[r]=x+1; r++;           }    for(int i=0;i<r;i++)       printf("%d\n",e[i]);    return 0;}