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[5010];
struct stick
{
    int l,w;
}sti[5010];
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=0;i<n;i++)
        scanf("%d%d",&sti[i].l,&sti[i].w);
        sort(sti,sti+n,cmp);
        int k=0;
        a[k]=sti[0].w;
        for(int i=1;i<n;i++)
        {
            int tem=10000000,temj;
            for(int j=0;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!=10000000)
            a[temj]=sti[i].w;
            else
            a[++k]=sti[i].w;
        }
        cout<<k+1<<endl;
    }
    return 0;
}