1051 Wooden Sticks

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12443    Accepted Submission(s): 5167


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
 
 1 #include<iostream>
 2 #include<math.h>
 3 #include<stdio.h>
 4 #include<stdlib.h>
 5 #include<string.h>
 6 #include<algorithm>
 7 using namespace std;
 8 struct ln{
 9     int x;
10     int y;
11     int z;
12 }a[5009];
13 bool cmp(const ln a,const ln b)
14 {
15     if(a.y==b.y)
16     return a.x<b.x;
17     else
18     return a.y<b.y;
19 }
20 int main()
21 {
22     //freopen("in.txt","r",stdin);
23     int t;
24     scanf("%d",&t);
25     while(t--)
26     {
27         memset(a,0,sizeof(a));
28         int n;
29         scanf("%d",&n);
30         for(int i=0;i<n;i++)
31         {
32             scanf("%d%d",&a[i].x,&a[i].y);
33             a[i].z=0;
34         }
35         sort(a,a+n,cmp);
36         int count=0,length;
37         for(int i=0;i<n;i++)
38         {
39             if(!a[i].z)
40             {
41                 count++;
42                 length=a[i].x;
43                 for(int j=i+1;j<n;j++)
44                 {
45                     if(!a[j].z)
46                     {
47                        if(a[j].x>=length)
48                        {
49                            a[j].z=1;
50                            length=a[j].x;
51                        }
52                     }
53 
54                 }
55             }
56 
57         }
58         printf("%d
",count);
59     }
60     return 0;
61 }
View Code
 
原文地址:https://www.cnblogs.com/xuesen1995/p/4105848.html