# Wooden Sticks

By | 2013/04/13

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

-2e9这个没看明白为什么是这个数，不过喜欢这个代码。下面是参考了网上的结果

```#include <stdio.h>
#include <string.h>
#include <stdlib.h>
struct z{
int x;
int y;
};
int comp(const void *a,const void *b) {
struct z *c=(struct z*)a;
struct z *d=(struct z*)b;
if(c->x==d->x)
return c->y-d->y;
return c->x-d->x;
}
int main() {
int t,a,b,m,n,i;
struct z g;
scanf("%d",&t);
while(t--) {
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%d%d",&g[i].x,&g[i].y);
qsort(g+1,n,sizeof(g),comp);
a=b=0;
while(b!=n){
m=-2e9;
for(i=1;i<=n;i++) {
if(g[i].y>=m) {
m=g[i].y;
g[i].y=-2e9-1;
b++;
}
}
a++;
}
printf("%d\n",a);
}
return 0;
}```