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
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
题意描述：
共有n个木棒，已知每个木棒的长度和重量，第一个木棒加工安装需要一分钟，在加工长度l和重量w的棒材之后，如果l<=l'和w<=w'，机器将不需要长度l'和重量w'的棒材的安装时间。否则，安装需要1分钟。求加工n个木棒的最小时间。
解题思路：
将木棒根据长度进行从小到大排序，找排序后木棒的重量序列的上升序列有多少段，就需要多少分钟。
程序代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct A
{
int a;
int b;
}q[5010];
int cmp(A x,A y)
{
if(x.a==y.a)
return x.b<y.b;
return x.a<y.a;
}
int main()
{
int t,n,i,j,num,ans,m,x,y;
int book[5010];
scanf("%d",&t);
while(t--)
{
memset(book,0,sizeof(book));
num=0;
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d%d",&q[i].a,&q[i].b);
sort(q,q+n,cmp);
ans=0;
while(ans<n)
{
for(i=0;i<n;i++)
{
if(book[i]==0)
{
book[i]=1;
x=q[i].a;
y=q[i].b;
ans++;
m=i;
num++;
break;
}
}
for(i=m;i<n;i++)
{
if(x<=q[i].a&&y<=q[i].b&&book[i]==0)
{
book[i]=1;
x=q[i].a;
y=q[i].b;
ans++;
}
}
}
printf("%d\n",num);
}
return 0;
}