2013
11-09

# Wooden Sticks

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 ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .

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 2n 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.

The output should contain the minimum setup time in minutes, one per line.

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


2
1
3

import java.util.*;

public class Main{
public static void main(String[] args){
int l[]=new int[10000];
int w[]=new int[10000];
int tt[]=new int[5000];
int k=0;
Scanner keyin=new Scanner(System.in);
int m=keyin.nextInt();
int mm=m;
int n=0;
for(;m>0;m--){
n=keyin.nextInt();
for(int j=0;j< n;j++){
l[j]=keyin.nextInt();
w[j]=keyin.nextInt();
}

int time=0,c=0;
for(int t=0;t<=n-2;t++){
for(int s=n-2;s>=t;s--){
if(l[s]>l[s+1]){
int temp=l[s];
l[s]=l[s+1];
l[s+1]=temp;
temp=w[s];
w[s]=w[s+1];
w[s+1]=temp;
}
}
}

for(int p=0;p< n;p++){
if(w[p]!=-1){
time++;
c=w[p];
for(int q=p+1;q< n;q++){
if(w[q]>=c){
c=w[q];
w[q]=-1;
}
}
}
}
tt[k++]=time;
}
for(int r=0;r< mm;r++)
System.out.println(tt[r]);
}
}

1. int half(int *array,int len,int key)
{
int l=0,r=len;
while(l<r)
{
int m=(l+r)>>1;
if(key>array )l=m+1;
else if(key<array )r=m;
else return m;
}
return -1;
}
这种就能避免一些Bug
l,m,r
左边是l,m;右边就是m+1,r;