HDU1051 贪心

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18745    Accepted Submission(s): 7692

Problem 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

 

Source

Asia 2001, Taejon (South Korea)

 

题意：

 有一个木棒处理机，每次处理一个木棒，处理的一个木棒时需要1分钟step时间，处理完一个木棒后下一次只能处理l和w都不小于此木棒的木棒，问最少的step时间。

代码：

 /*
 按照l由小到大排序，用一个数组存储机器不需要step的w值，如果下一个w大于等于此数组中的某一个用它更新与他差值
 最小的那个，如果大于每一个数组值就再开一个数组来记录他。
 */
 #include<iostream>
 #include<string>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<vector>
 #include<iomanip>
 #include<queue>
 #include<stack>
 using namespace std;
 int t,n;
 int a[];
 struct stick
 {
     int l,w;
 }sti[];
 bool cmp(stick x,stick y)
 {
     if(x.l==y.l)
     return x.w<y.w;
     else return x.l<y.l;
 }
 int main()
 {
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&n);
         for(int i=;i<n;i++)
         scanf("%d%d",&sti[i].l,&sti[i].w);
         sort(sti,sti+n,cmp);
         int k=;
         a[k]=sti[].w;
         for(int i=;i<n;i++)
         {
             int tem=,temj;
             for(int j=;j<=k;j++)
             {
                 if((sti[i].w>=a[j])&&(sti[i].w-a[j]<tem))
                 {
                     tem=sti[i].w-a[j];
                     temj=j;
                 }
             }
             if(tem!=)
             a[temj]=sti[i].w;
             else
             a[++k]=sti[i].w;
         }
         cout<<k+<<endl;
     }
     return ;
 }