Wooden Sticks

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

 

翻译一下题目：

现有n根木棒，已知它们的长度和重量。要用一部木工机一根一根地加工这些木棒。该机器在加工过程中需要一定的准备时间，是用于清洗机器，调整工具和模板的。木工机需要的准备时间如下：
（1）第一根木棒需要1min的准备时间；
（2）在加工了一根长为l，重为w的木棒之后，接着加工一根长为ll（l<=ll），重为ww（w<=ww）的木棒是不需要任何准备时间的。否则需要一分钟的准备时间。
给定n根木棒，你要找到最少的准备时间。例如现在有长和重分别为（4,9），（5,2），（2,1），（3,5）和（1,4）的五根木棒，那么所需准备时间最少为2min，顺序为（1,4），（3,5），（4,9），（2,1），（5,2）。

 

简单的贪心，只需要不停的扫描这n个木棍，符合条件就继续往后扫，已经比过的直接跳过，到最后没了时间就+1，然后再从头扫描，用一个cnt来计数，循环结束的条件是cnt==n。  注意一定要先排序！！

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>       //sort函数所需头文件
using namespace std;

struct stick
{
    int l,w;       //长度和重量
    int v;        //用于标记木棒是否被处理
}s[5500];

int cmp(const stick &a,const stick &b)
{
    if(a.l==b.l)
        return a.w<b.w;
    return a.l<b.l;
}              //定义排序原则

int main()
{
    int i,t,k,n,cnt,ans;       //cnt表示被标记元素的个数（如果元素全部被标记cnt=n说明他们都找到了其升序子序列），ans为准备时间
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d%d",&s[i].l,&s[i].w);
            s[i].v=0;     //初始标记为0
        }
        sort(s,s+n,cmp);
        cnt=0;
        k=0;
        ans=0;
        while(cnt!=n)
        {
            k=0;
            while(k<n&&s[k].v==1)     //依次遍历到上次循环找到的升序子序列 直到不满足条件的第一个元素
                k++;
            ans++;           //准备时间要+1
            for(i=k;i<n;i++)           //找出一个从k开始的升序子序列 并标记为1
            {
                if(s[i].v==0&&s[i].l>=s[k].l&&s[i].w>=s[k].w)   //之前找过的已经标记为1了，所以不可以再次使用
                {
                    s[i].v=1;
                    cnt++;    //记录标记元素的个数
                    k=i;      //k指向当前处理的木棒
                }
            }
        }
        printf("%d
",ans);
    }
    return 0;
}