Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18745    Accepted Submission(s): 7692

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

Source

Asia 2001, Taejon (South Korea)

题意：

有一个木棒处理机，每次处理一个木棒，处理的一个木棒时需要1分钟step时间，处理完一个木棒后下一次只能处理l和w都不小于此木棒的木棒，问最少的step时间。

代码：

 1 /*
 2 按照l由小到大排序，用一个数组存储机器不需要step的w值，如果下一个w大于等于此数组中的某一个用它更新与他差值
 3 最小的那个，如果大于每一个数组值就再开一个数组来记录他。
 4 */
 5 #include<iostream>
 6 #include<string>
 7 #include<cstdio>
 8 #include<cmath>
 9 #include<cstring>
10 #include<algorithm>
11 #include<vector>
12 #include<iomanip>
13 #include<queue>
14 #include<stack>
15 using namespace std;
16 int t,n;
17 int a[5010];
18 struct stick
19 {
20     int l,w;
21 }sti[5010];
22 bool cmp(stick x,stick y)
23 {
24     if(x.l==y.l)
25     return x.w<y.w;
26     else return x.l<y.l;
27 }
28 int main()
29 {
30     scanf("%d",&t);
31     while(t--)
32     {
33         scanf("%d",&n);
34         for(int i=0;i<n;i++)
35         scanf("%d%d",&sti[i].l,&sti[i].w);
36         sort(sti,sti+n,cmp);
37         int k=0;
38         a[k]=sti[0].w;
39         for(int i=1;i<n;i++)
40         {
41             int tem=10000000,temj;
42             for(int j=0;j<=k;j++)
43             {
44                 if((sti[i].w>=a[j])&&(sti[i].w-a[j]<tem))
45                 {
46                     tem=sti[i].w-a[j];
47                     temj=j;
48                 }
49             }
50             if(tem!=10000000)
51             a[temj]=sti[i].w;
52             else
53             a[++k]=sti[i].w;
54         }
55         cout<<k+1<<endl;
56     }
57     return 0;
58 }