1048: Wooden Sticks
| Result | TIME Limit | MEMORY Limit | Run Times | AC Times | JUDGE |
|---|---|---|---|---|---|
| 3s | 8192K | 884 | 291 | Standard |
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
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#include<algorithm>
#include<string.h>
using namespace std;
class NODE
{
public:
int x;int y;
};
int cmp( const NODE &a, const NODE &b ){
if( a.x < b.x )
return 1;
else
if( a.x == b.x ){
if( a.y < b.y )
return 1;
else
return 0;
}
else
return 0;
}
int main()
{
freopen("in.txt","r",stdin);
NODE node[5005];
int visited[5005];
int t,n;
scanf("%d",&t);
for(;t>0;t--)
{
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d%d",&node[i].x,&node[i].y);
visited[i]=0;
}
int m=0;
sort(node,node+n,cmp);
/*for(int i=0;i<n;i++)
{
printf("%d %d\n",node[i].x,node[i].y);
}
printf("\n\n\n\n\n");*/
for(int i=0;i<n;i++)
{
if(!visited[i])
{
m++;
visited[i]=1;
int y=node[i].y;
for(int j=i+1;j<n;j++)
{
if(node[j].y>=y&&!visited[j])
{
visited[j]=1;
y=node[j].y;
}
}
}
}
printf("%d\n",m);
}
return 0;
}