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);

     }

 }

巴特西