贪心算法 Wooden Sticks

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



9 5 2 2 1 3 5 1 4 

2 1 1 2 2 

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。

posted on 2013-05-26 01:22 straw_berry 阅读(333) 评论(0) 编辑收藏举报