Problem B

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

 简单题意：

  有一堆木棍需要处理，事先已经知道了他们的长度和重量。木棍处理机器需要为第一根木棍留下1min的设置时间，然后其他的木棍如果长度比第一根大并且重量也比第一根重，那么机器将不需要设置时间。否则，需要重新开始设置，耗费1min的设置时间。编写一个程序，使得处理完这批木棍的时间最短。

解题思路形成过程：

  这是典型的贪心算法，并不难。和Problem A一样，我拿到题目就想，分批次完成，使用迭代。虽然Problem A没成功，但是思想肯定没有错误。

感想：

  对于这个题，我想说的太多了。ACM不只是考思路，像这个题目，我有两种AC 代码。但是开始一直Runtime Error，最后查明白了是数组界限的问题。提交20次，这个经验教训，我想，很难忘记了。然后就是Time limit Exceeded，真是怎么提交怎么不对。最后，是输入，不能使用while(cin>>n)，题目也说的不是很明白，只是测试一组数据。下面附上两种AC代码。

AC代码：

 

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
struct stick
{
    int l;
    int w;
};
bool cmp(const stick &a,const stick &b)
{
    if(a.l<=b.l)return true;//所有的数据从小到大排序
    return false;
    }
void greedy_select(int n,stick a[],bool b[])
{
    b[1]=true;
    int preEnd=1;
    for(int i=2;i<=n;i++)
    {
        if(a[i].l>=a[preEnd].l&&a[i].w>=a[preEnd].w)
        {
            b[i]=true;
            preEnd=i;
        }
        else b[i]=false;
        }
    }
int main()
{
    int n,stickNum;
    stick s[5001];
    bool f[5001];
    scanf("%d",&n);
    {
        for(int k=0;k<n;k++)
        {
        scanf("%d",&stickNum);
        for(int i=1;i<=stickNum;i++)
        {
            scanf("%d %d",&s[i].l,&s[i].w);
            }
        sort(s+1,s+stickNum+1,cmp);
        int sum=0;
        while(stickNum!=0)
        {
            greedy_select(stickNum,s,f);
            int j=1,total=0;
            for(int i=1;i<=stickNum;i++)
            {
            if(f[i]==true){total++;}
            else
            {
                s[j]=s[i];
                //cout<<"j:"<<j<<endl;
                j++;
                }
            }
            //cout<<"total:"<<total<<endl;
            sum++;
            stickNum-=total;
        }

        printf("%d\n",sum);
        }
    }
    return 0;
}

第二种：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
struct stick
{
    int l;
    int w;
};
bool cmp(const stick &a,const stick &b)
{
    if(a.l<=b.l)return true;//所有的数据从小到大排序
    return false;
    }
void greedy_select(int n,stick a[],bool b[])
{
    b[1]=true;
    int preEnd=1;
    for(int i=2;i<=n;i++)
    {
        if(a[i].w<a[preEnd].w)
        {
            b[i]=true;
            preEnd=i;
        }
        else b[i]=false;
        }
    }
int main()
{
    int n,stickNum;
    stick s[5001];
    bool f[5001];
    scanf("%d",&n)；
    {
        for(int k=0;k<n;k++)
        {
        scanf("%d",&stickNum);
        for(int i=1;i<=stickNum;i++)
        {
            scanf("%d %d",&s[i].l,&s[i].w);
            }
        sort(s+1,s+stickNum+1,cmp);
            greedy_select(stickNum,s,f);
            int total=0;
            for(int i=1;i<=stickNum;i++)
            if(f[i])total++;
        printf("%d\n",total);
        }
    }
    return 0;
}