1051 Wooden Sticks（贪心-3）

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>
#include <cstring>
#include <stdio.h>
#include <math.h>
using namespace std;
struct sa
{
    int x;
    int y;
    int flag;
}data[10005];
int cmp(const sa &a,const sa &b)
{
    if(a.x!=b.x)
    return a.x<b.x;
    else
    return a.y<b.y;
}
int main()
{
    int n,m,ans,tmp1,tmp2;
    while(cin>>n)
    {
        while(n--)
        {
            cin>>m;
            for(int i=0;i<m;i++)
            {
                cin>>data[i].x>>data[i].y;
                data[i].flag=0;
            }
            sort(data,data+m,cmp);
            ans=0;
            for(int i=0;i<m;i++)
            {
                if(data[i].flag!=1)
                {
                    ans++;
                    tmp1=data[i].x;tmp2=data[i].y;data[i].flag=1;
                    for(int j=i+1;j<m;j++)
                    if(data[j].x>=tmp1&&data[j].y>=tmp2&&data[j].flag==0)
                    {
                        tmp1=data[j].x;tmp2=data[j].y;data[j].flag=1;
                    }
                }
            }
            cout<<ans<<endl;
        }
    }
    return 0;
}

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