Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13836    Accepted Submission(s): 5716

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

213

 

Source

Asia 2001, Taejon (South Korea)

#include <iostream>#include<algorithm>using namespace std;struct ac{    int l,w;    int flag;}node[5005];bool cmp(ac a,ac b){     if(a.l<b.l)        return true;     else if(a.l>b.l)         return false;     else       return a.w<b.w;}int main(){    int n,t,i,j,count,k;    cin>>t;    while(t--)    {        count=0;        for(i=0;i<5005;i++)            node[i].flag=0;        cin>>n;        for(i=0;i<n;i++)        {            cin>>node[i].l;            cin>>node[i].w;        }        sort(node,node+n,cmp);        for(i=0;i<n;i++)        {            if(node[i].flag==1)                continue;            count++;            k=i;            node[i].flag=1;            for(j=i+1;j<n;j++)            {                if(node[j].l>=node[k].l&&node[j].w>=node[k].w&&node[j].flag==0)                {                    node[j].flag=1;                    k=j;                }            }        }        cout<<count<<endl;    }    return 0;}