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# Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16247    Accepted Submission(s): 6668

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

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## 迷失在幽谷中的鸟儿，独自飞翔在这偌大的天地间，却不知自己该飞往何方……

```#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
struct mood
{
int w;
int l;
} d;
bool cmp(mood a,mood b)
{
if(a.l==b.l)
return a.w<b.w;
return a.l<b.l;
}
int k;
int main()
{
int n;
cin>>n;
while(n--)
{
int m;
cin>>m;
for(int i=0; i<m; i++)
cin>>d[i].l>>d[i].w;
sort(d,d+m,cmp);
memset(k,-1,sizeof(k));
int flag=0;
for(int i=0; i<m; i++)
for(int j=0; j<=m; j++)
if(d[i].w>=k[j])
{
k[j]=d[i].w;
if(j+1>flag)flag=j+1;
break;
}
cout<<flag<<endl;
}
return 0;
}
```

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• 原文地址：https://www.cnblogs.com/im0qianqian/p/5989549.html