Poj1065 wooden sticks (Greedy + Dynamic Planning-monotonic incrementing subsequence)

 

Description

C. Has some wood sticks, whose length and quality are known. A machine is required to handle these sticks, which takes a unit of time to start, if the weight and length of the I + 1 wood rod are greater than or equal

 

The I-th processed wooden stick will not take time, otherwise it will take a unit of time. C Xiaojia wants to finish processing the wooden stick in the shortest time because he is in a hurry to go on a date. Can you tell him what to do?

 

Input

The first line is an integer T, indicating a total of T groups of input data.

The first line of each group of test data is an integer N (1 <= n <= 5000), indicating that there are n sticks. In the next line, enter the L, w (0 <L, W <= 10000) of N wooden bars, and separate them with a space, indicating

 

Length and quality of wood sticks.

 

Output

The shortest time for processing these sticks.

 

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

 

 

The Code is as follows:

 

# Include <iostream> <br/> # include <algorithm> <br/> # include <string. h> <br/> using namespace STD; <br/> struct machine <br/>{< br/> int L, W; <br/>} data [5010]; <br/> bool use [5010]; // indicates whether the wooden rod has been used: 1 used, 0 used <br/> bool operator <(const machine, const machine B) // sort the sticks in ascending order. If they are equal, the weights increase. <br/>{< br/> If (. L = B. l) <br/> return. W <B. w; <br/> return. L <B. l; <br/>}< br/> int main () <br/> {<br/> // freopen ("1.txt"," R ", Stdin); <br/> int t; <br/> CIN> T; <br/> while (t --) <br/>{< br/> memset (use, 0, sizeof (use); <br/> int N, Count = 0; <br/> machine last; <br/> CIN> N; <br/> for (INT I = 0; I <n; ++ I) <br/>{< br/> CIN> data [I]. l> data [I]. w; <br/>}< br/> sort (data, data + n); // wood rod sorting <br/> for (INT I = 0; I <N; + + I) // returns the ascending subsequence of the I-th wooden stick. <br/>{< br/> last. W = data [I]. w; // record the start of the new sequence <br/> If (! Use [I]) <br/>{< br/> for (Int J = I + 1; j <n; j ++) <br/>{< br/> If (last. W <= data [J]. W &&! Use [J]) // L No comparison is required for sorting <br/>{< br/> use [J] = 1; // when used, mark it as 1 <br/> last. W = data [J]. w; // team end element <br/>}< br/> count ++; // after the sub-sequence ends, the time is incremented by 1. calculate the I + 1 wooden rod <br/>}< br/> cout <count <Endl; <br/>}< br/> return 0; <br/>} 

 

Solution: (Greedy Algorithm + Dynamic Planning-monotonic incrementing subsequence)

 

1. First, sort the wooden sticks. The overall order is l incremental. If l is equal, W increments.

 

2. Processing the I-th wooden stick: selects the monotonic incrementing subsequence and marks the monotonic incrementing subsequence as 1 (used ). After selection, the time is + 1. Processing of the I + 1 wooden rod