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 #include<stdio.h>
2 #include<stdlib.h>
3 #include<
string.h>
4 #define N 5000;
5
6 struct node
7 {
8 int l;
9 int w;
10 int flag;
11 }sticks[
5000];
12 int cmp(
const void *p,
const void *
q)
13 {
14 struct node *a = (
struct node *
)p;
15 struct node *b = (
struct node *
)q;
16 if(a->l > b->l)
return 1;
17 else if(a->l < b->l)
return -
1;
18 else return a->w > b->w ?
1 : -
1;
19 }
20 int main()
21 {
22 int t,n,cnt,cl,cw;
23 int i,j;
24 scanf(
"%d",&
t);
25 while(t--
)
26 {
27 memset(sticks,
0,
sizeof(sticks));
28 scanf(
"%d",&
n);
29 for( i =
0; i < n; i++
)
30 scanf(
"%d %d",&sticks[i].l,&
sticks[i].w);
31 qsort(sticks,n,
sizeof(sticks[
0]),cmp);
32 sticks[
0].flag =
1;
33 cl = sticks[
0].l;
34 cw = sticks[
0].w;
35 cnt =
1;
36 for( j =
1; j < n; j++
)
37 {
38 for( i = j; i < n; i++
)
39 {
40 if(!sticks[i].flag&&sticks[i].l>=cl&&sticks[i].w>=
cw)
41 {
42 cl =
sticks[i].l;
43 cw =
sticks[i].w;
44 sticks[i].flag =
1;
45 }
46 }
47 i =
1;
48 while(sticks[i].flag)
49 i++
;
50 j =
i;
51 if(j == n)
break;
52 cl =
sticks[j].l;
53 cw =
sticks[j].w;
54 sticks[j].flag =
1;
55 cnt++
;
56 }
57 printf(
"%d\n",cnt);
58
59 }
60 return 0;
61 }
题意：
我们要处理一些木棍，第一根的时间是1分钟，另外的，在长度为l，重为w的木棍后面的那根的长度为l’, 重量w’，只要l <=l’ 并且w <= w’，就不需要时间，否则需要1分钟，求如何安排处理木棍的顺序，才能使花的时间最少。
思路：
贪心算法。先把这些木棍按长度和重量都从小到大的顺序排列。cl和cw是第一根的长度和重量，依次比较后面的是不是比当前的cl，cw大，是的话把标志flag设为1，并跟新cl,cw。比较完后，再从前往后扫描，找到第一个标志位为0的，作为是第二批的最小的一根，计数器加一。把它的长度和重量作为当前的cl，cw，再循环往后比较。直到所有的都处理了。
转载于:https://www.cnblogs.com/LK1994/archive/2013/05/26/3099538.html
