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。
1 #include<cstdio>
2 #include<algorithm>
3 using namespace std;
4 struct stu
5 {
6 int l,w;
7 }st[5010];
8 bool cmp(stu a,stu b)
9 {
10 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
11 return a.l<b.l;
12 else
13 return a.w<b.w;
14 }
15
16 int main()
17 {
18 int t;
19 scanf("%d",&t);
20 while(t--)
21 {
22 int i,j,d[5010];
23 int n;
24 scanf("%d",&n);
25 for(i = 0 ; i < n ; i++)
26 {
27 scanf("%d %d",&st[i].l,&st[i].w);
28 d[i]=0; //标记查询过的木头，0为未标记
29 }
30 int ans=0;
31 sort(st,st+n,cmp);
32 for(i = 0 ; i < n ; i++)
33 {
34 if(!d[i])
35 {
36 d[i]=1;
37 ans++; //ans表示时间
38 int w=st[i].w;
39 for(j = i+1 ; j < n ; j++) //查询i以后的木头
40 {
41 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
42 {
43 d[j]=1; //标记在i后处理不用时间的木头
44 w=st[j].w;
45 }
46
47 }
48 }
49 }
50 printf("%d\n",ans);
51 }
52 }