hdu1051 Wooden Sticks(贪心+排序，逻辑)

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26095 Accepted Submission(s): 10554

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

4 9 5 2 2 1 3 5 1 4

2 2 1 1 2 2

1 3 2 2 3 1

Sample Output

有若干组木棍：木棍的长度为I 重为W 现在用机器加工这些木头：规则如下：

1；加工第一组木头会花去1分钟
2：设加工的下一组木头的长度为 I' 重量为W‘ 如果 I=<I' 且 W=<W' ,则加工这根木头不需要花费时间。

问，所花费的最少时间是多少？

可以先按长度和重量从小到大排序，长度相同按重量从小到大排

每次从最前面没选过的开始，计算有多少种递增子序列，加一点技巧会更清楚，具体看代码中间的while部分

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 struct node
 4 {
 5     int l,w,flag;
 6 }a[5005];
 7 void init()
 8 {
 9     for(int i=0;i<5005;i++)
10     {
11         a[i].l=0;a[i].w=0;a[i].flag=1;
12     }
13 }
14 bool cmp(node x,node y)
15 {
16     if(x.l==y.l)return x.w<y.w;
17     else return x.l<y.l;
18 }
19 int main()
20 {
21     int t;
22     while(~scanf("%d",&t))
23     {
24         while(t--)
25         {
26             init();
27             int n;
28             scanf("%d",&n);
29             for(int i=0;i<n;i++)
30             {
31                 scanf("%d %d",&a[i].l,&a[i].w);
32             }
33             sort(a,a+n,cmp);
34 //            for(int i=0;i<n;i++)
35 //            {
36 //                printf("%d %d
",a[i].l,a[i].w);
37 //            }
38             int ans=0,j=0;
39             while(j<n)//判断有没有全部选完 ，核心部分 
40             {
41                 ans++;
42                 int tempx=0,tempy=0;
43                 for(int i=0;i<n;i++)
44                 {
45                     if(a[i].l>=tempx&&a[i].w>=tempy&&a[i].flag)//flag==1表示没有被选过 
46                     {
47                         a[i].flag=0;//选过的标记，下次就不选了 
48                         j++;
49                         tempx=a[i].l;
50                         tempy=a[i].w;
51                     }
52                 }    
53             }
54             printf("%d
",ans);
55         }
56     }
57     return 0;
58 }