Wooden Sticks---（贪心）

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

 

Source

Asia 2001, Taejon (South Korea)

分析：贪心策略，将pair按照first进行升序排序，按照不减原则选择重量，选一个，消灭一个。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 5000+10;
#define INF 0x3f3f3f3f
int t;
pair<int,int> arr[maxn];
bool cmp(pair<int,int> a,pair<int,int> b){
    if(a.first<b.first) return true;
    else if(a.first==b.first) return a.second-b.second;
    else return false;
}

int main(){
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        for( int i=0; i<n; i++ ){
            scanf("%d%d",&arr[i].first,&arr[i].second);
        }
        sort(arr,arr+n,cmp);
        /*或者直接sort(arr,arr+n);*/
        int ans=0;
        for( int i=0; i<n; i++ ){
            if(arr[i].second==-1) continue;
            int weg=arr[i].second;
            ans++;
            for( int j=i+1; j<n; j++ ){
                if(arr[j].second>=weg){
                    weg=arr[j].second;
                    arr[j].second=-1;
                }
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}