Wooden Sticks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6563 Accepted Submission(s): 2707
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
1 // 贪心算法---先排序---后选择第一个没有用过的木头一次向后找,用掉所有可以用掉的木头,然后返回第一个没用过的木头继续找
2 #include<cstdio>
3 #include<algorithm>
4 using namespace std;
5 struct stick
6 {
7 int len;
8 int wei;
9 }a[5001];
10 bool used[5001];
11
12 bool cmp(stick k1,stick k2) //按照长度从小到大排序,若长度相同按照重量递增排序
13 {
14 if(k1.len==k2.len)
15 return k1.wei<k2.wei;
16 else
17 return k1.len<k2.len;
18 }
19
20 int main()
21 {
22 int T,n,i,j,st,count;
23 bool flag;
24 scanf("%d",&T);
25 while(T--)
26 {
27 scanf("%d",&n);
28 for(i=0;i<n;++i)
29 scanf("%d%d",&a[i].len,&a[i].wei);
30 sort(a,a+n,cmp);
31 memset(used,false,sizeof(used));
32 used[0]=true;
33 st=0; //记录第一个没有用过的木头
34 count=0;
35 while(st<n)
36 {
37 ++count;
38 for(i=st+1,j=st,flag=true;i<n;++i)
39 {
40 if(used[i])
41 continue;
42 if(a[j].len<=a[i].len&&a[j].wei<=a[i].wei)
43 {
44 used[i]=true;
45 j=i;
46 }
47 else
48 {
49 if(flag)
50 {
51 st=i; //只记录第一个没用过的木头
52 flag=false;
53 }
54 }
55 }
56 if(flag) //说明都用过了
57 break;
58 }
59 printf("%d\n",count);
60 }
61 return 0;
62 }