C-wooden Sticks

**Time ****Limit:**1000MS

**Memory Limit:**32768KB

**64bit IO Format:**%I64D &%i64u Submit Status

Description There is a pile of n wooden sticks. The length and weight of each stick is known in advance. The sticks is 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 is associated with cleaning operations and changing tools and shapes. The setup times of the woodworking machine is 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 would need no setup time for a stick of length l ' and weight W ' if l<=l ' and W<=w '. Otherwise, it'll need 1 minute for setup.

You is 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 is (4,9), (5,2), (2,1), (3,5), and (1,4), then the Minimum setup time should be 2 minutes since there are 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 on the first line of the input file. Each test case consists of a lines:the first line has a integer n, 1<=n<=5000, that represents the number of Wo Oden sticks in the test case, and the second line contains n 2 positive integers L1, W1, L2, W2, ..., LN, WN, each of Magn Itude at most 10000, where Li and wi is the length and weight of the i th wooden stick, respectively. The 2n integers is 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 This question is the train of thought that the study elder sister gives. is to sort by length before using flag to record whether or not it has been accessed.

#include <iostream>
#include <algorithm>
using namespace std;
typedef struct dot{
int l,w;
bool Flag;
} Dot;
BOOL CMP (dot A,dot b)
{
if (A.L!=A.L) return a.w<a.w;
else return a.l<b.l;
}
int main ()
{
int t,n;
cin>>t;
while (t--)
{
cin>>n;
Dot a[10000];
for (int i=0;i!=n;i++) {
cin>>a[i].l>>a[i].w;
a[i].flag=0;
}
Sort (a,a+n,cmp);
int sum=0;
for (int i=0;i!=n;i++) {
if (a[i].flag==0) {
sum++;
a[i].flag=1;
int k=i;
for (int j=k;j!=n;j++) {
if (!A[J].FLAG&&A[K].W<=A[J].W) {k=j;a[j].flag=1;
}}}} cout<<sum<<endl;
}
return 0;
}

The greedy practice is not good ....