杭电 1051 Wooden Sticks

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

大意：
　　

描述

C小加有一些木棒，它们的长度和质量都已经知道，需要一个机器处理这些木棒，机器开启的时候需要耗费一个单位的时间，如果第i+1个木棒的重量和长度都大于等于

第i个处理的木棒，那么将不会耗费时间，否则需要消耗一个单位的时间。因为急着去约会，C小加想在最短的时间内把木棒处理完，你能告诉他应该怎样做吗？

输入

第一行是一个整数T，表示输入数据一共有T组。

每组测试数据的第一行是一个整数N（1<=N<=5000）,表示有N个木棒。接下来的一行分别输入N个木棒的L，W（0 < L ,W <= 10000），用一个空格隔开，分别表示

木棒的长度和质量。

输出

处理这些木棒的最短时间。

先将木头按长度从小到大排列，然后假设从第一个木头开始，查询第2~n个木头，把不用时间的全部标记，再找到下一个未标记的木头为开始，时间加一，再将这块木头后面不用时间的全部标记，以此类推。

 #include<cstdio>
 #include<algorithm>
 using namespace std;
 struct stu
 {
     int l,w;
 }st[];
 bool cmp(stu a,stu b)
 {
     if(a.l != b.l)                //将木头按长度从小到大排列，长度一样按质量从小到大排列
         return a.l<b.l;
     else
         return a.w<b.w;
 }

 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--)
     {
         int i,j,d[];
         int n;
         scanf("%d",&n);
         for(i =  ; i < n ; i++)
         {
             scanf("%d %d",&st[i].l,&st[i].w);
             d[i]=;                                //标记查询过的木头，0为未标记
         }
         int ans=;
         sort(st,st+n,cmp);
         for(i =  ; i < n ; i++)
         {
             if(!d[i])
             {
                 d[i]=;
                 ans++;                            //ans表示时间
                 int w=st[i].w;
                 for(j = i+ ; j < n ; j++)        //查询i以后的木头
                 {
                     if(!d[j] && st[j].w >= w)    //因为木头的长度是从小到大排列，所以这只判断重量
                     {
                         d[j]=;                    //标记在i后处理不用时间的木头
                         w=st[j].w;
                     }

                 }
             }
         }
         printf("%d\n",ans);
     }
 }