贪心，求最短的时间，将可以省去的数字用0代替然后后来碰见0就continue（链表做会更节约时间）

Problem L

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 29 Accepted Submission(s) : 8

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



Output for the Sample Input

2

1

3

代码：

#include<iostream>

using namespace std;

#include<algorithm>

struct stick

{int x,y;

}s[5005];

int cmp(stick a,stick b)

{

if(a.x==b.x)return a.y<=b.y; //在x相等的时候对y排一下序

else return a.x<b.x; //x不相等的时候对x排序

}

int main()

{

int n,m,i,max,t,k;

cin>>k;

while(k--)

{

cin>>n;

for(i=0;i<n;i++)

cin>>s[i].x>>s[i].y;

m=n;

sort(s,s+m,cmp); //用sort排一下序，先将x排序，在x的基础上来对y操作（排序）

t=0;

while(m!=0)

{

for(i=0,max=-1;i<n;i++)

{

if(s[i].y==0)continue;

else if(s[i].y>=max)

{

max=s[i].y;

s[i].y=0; //将可以省去的数字用0代替，然后碰见0就continue

m--;

}

}

t++;

}

cout<<t<<endl;

}

return 0;

}