题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1051

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13534    Accepted Submission(s):
5590

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

Sample Output

2

1

3

题目大意：一堆木棍，有长度和重量，处理第一根需要一分钟，然后处理下一根木棍，如果这根木棍的长度和重量都不大于之前的木棍，则不需要花费时间，否则也需要一分钟。

分析：首先排序，先按长度排序，如果长度一样，则按重量排序（从小到大和从大到小排序都可以，我是按从小到大排序的）。然后求递增（减）序列的最小个数。

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <iostream>
 5 #include <algorithm>
 6 using namespace std;
 7
 8 int T,n;
 9 int a[5005],b[5005],vis[5001];
10
11 int main ()
12 {
13     int i,j;
14     int time;
15     int x,y;
16     scanf ("%d",&T);
17     while (T--)
18     {
19         scanf ("%d",&n);
20         for (i=0; i<n; i++)
21             scanf ("%d %d",&a[i], &b[i]);
22         for (i=0; i<n; i++)//两个for循环排序，没用结构体，比较麻烦
23         {
24             for (j=i; j<n; j++)
25             {
26                 if (a[i] > a[j])
27                 {
28                     swap(a[i], a[j]);
29                     swap(b[i], b[j]);
30                 }
31                 if (a[i] == a[j])
32                 {
33                     if (b[i] > b[j])
34                     {
35                         swap(a[i], a[j]);
36                         swap(b[i], b[j]);
37                     }
38                 }
39
40             }
41         }
42         time = 0;
43         memset(vis, 0, sizeof(vis));
44         for (i=0; i<n; i++)
45         {
46             if (vis[i])//若该木棍被访问过，跳过
47                 continue;
48             //x = a[i];
49             y = b[i];
50             for (j=i+1; j<n; j++)
51             {
52                 if (!vis[j] && b[j] >= y)
53                 {
54
55                     vis[j] = 1;//标记访问
56                     //x = a[j];
57                     y = b[j];//当前最大重量
58                 }
59             }
60             time++;//时间加一
61         }
62         printf ("%d\n",time);
63     }
64     return 0;
65 }