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).

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.

The output should contain the minimum setup time in minutes, one per line.

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

2
1
3

Asia 2001, Taejon (South Korea)

We have carefully selected several similar problems for you:  1052 1045 1009 1053 1789 

今天是端午节，祝各位C友端午节安康！好！废话不多说，开始我们的每日一水时间！

终于来了一道不是那么水的题了，首先解释一下题目的大意：

题目大意：

给你n根木棍的长度和重量。根据要求求出制作该木棍的最短时间。建立第一个木棍需要1分钟，如果接着制作的木棍比这个木棍的长度长，重量要重，那么接着制作的木棍不需要花费时间！然后如果再继续接着制作，则下一个木棍要比上一个木棍的长度长，重量大，则这个木棍也不需要花费时间！依次类推，反之，则需要花费一分钟，然后让你求出制作这一批木棍花费的最少的时间是多少！

解题思路：

其实大概明白了题意，便可以知道，这题考的是贪心！然后关键就在于贪心策略的选择！题目说了求最小时间，那就肯定要求长度和重量都要相对来说最小，我的策略如下：

1.首先按照木棍的长度进行排序，如果长度相等则重量重的放在前面。

2.当然排序完后，不一定都是下一根木棍重量和长度都大于前一根的，但是我们可以根据排序建立的数组再进行多次扫描，比较。

3.根据建立的数组依次比较他们的重量，如果大于标记数组跳过，否则ans加1。

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 5010;
struct stick
{
    int length;
    int weight;
    bool vis;
};
bool cmp(stick s1,stick s2)
{
    if(s1.length==s2.length)
        return s1.weight<s2.weight;
    else
        return s1.length<s2.length;
}
int n;
stick s[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&s[i].length,&s[i].weight);
            s[i].vis = false;
        }

        sort(s,s+n,cmp);
        int ans = 0;

        for(int i=0;i<n;i++)
        {
            if(!s[i].vis)
            {
                s[i].vis=true;
                ++ans;
                int weight = s[i].weight;
                for(int j=i+1;j<n;j++)
                {
                    if(!s[j].vis && s[j].weight>=weight)
                    {
                        s[j].vis = true;
                        weight = s[j].weight;
                    }
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}