Wooden Sticks_贪心

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21886		Accepted: 9341

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 ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .

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 2n 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<stdio.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>
#include<queue>
using namespace std;
struct node
{
    int l,w;
} a[5050];
bool operator <(node a,node b)
{
    if(a.l==b.l) return a.w<b.w;
    else return a.l<b.l;
}
int vis[5050],k;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        memset(vis,0,sizeof(vis));
        for(int i=1; i<=n; i++)
        {
            cin>>a[i].l>>a[i].w;
        }
        node tt;
        sort(a,a+n);
        k=0;
        for(int i=1; i<=n; i++)
        {
            tt.w=a[i].w;
            if(!vis[i])
            {
                for(int j=i+1; j<=n; j++)
                {
                      if(tt.w<=a[j].w&&!vis[j])
                      {
                          vis[j]=1;
                          tt.w=a[j].w;
                      }
                }
                k++;
            }
        }
        cout<<k<<endl;
    }
    return 0;
}