贪心-Wooden Sticks

先将火柴按照长度（或重量）优先排序，在不断遍历数组，找出其中重量（长度）递增子序列，并标记

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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>
using namespace std;
struct wood{
    int l;
    int w;
    int flag;
}W[];
bool cmp(wood w1, wood w2){
    if(w1.l<w2.l )
    return true;
    else if(w1.l==w2.l && w1.w<=w2.w)
    return true;
    return false;
}
int main()
{
    int t,n;
    cin>>t;
    while(t--){
        cin>>n;
        for(int i=;i<n;i++){
            cin>>W[i].l>>W[i].w;
            W[i].flag=;
        }
        sort(W,W+n,cmp);
        //for(int i=0;i<n;i++)
        //cout<<W[i].l<<" "<<W[i].w<<endl;
        int tw;
        int result=;
        for(int i=;i<n;i++){
            if(W[i].flag)
            continue;
            tw=W[i].w;
            W[i].flag=;
            for(int j=;j<n;j++){
                if(W[j].flag)
                continue;
                if(W[j].w>=tw){
                    tw=W[j].w;
                    W[j].flag=;
                }
            }
            result++;
        }
        cout<<result<<endl;
    }
    return ;
}