POJ1065--Wooden Sticks(动态规划)

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<algorithm> 
#include<string.h> 
using namespace std;

struct node{
    int l,w;
}stick[5005]; 
int use[5005];
bool operator<(node a,node b){
    if(a.l==b.l)
        return a.w<b.w;
    else
        return a.l<b.l;
}
int main(){
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        memset(use,0,sizeof(use));
        for(int i=0;i<n;i++){
            cin>>stick[i].l>>stick[i].w;
        }
        sort(stick,stick+n);
        int ans=0;
        node last;
        for(int i=0;i<n;i++){
            last.w=stick[i].w;
            if(use[i]==0){
                for(int j=i+1;j<n;j++){
                    if(last.w<=stick[j].w&&use[j]==0){
                        last.w=stick[j].w;
                        use[j]=1;
                    }
                }
                ans++;
            } 
        }
        cout<<ans<<endl;
    }
    return 0;
}