HDU 1051 Wooden Sticks

Posted 2021-01-08 shenyuling

 

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25661    Accepted Submission(s): 10401

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

 

 

 

思路：1 .先按长度从小到大排序(如果长度相同，在比较weight) 

           2. 长度搞好了以后，现在只需看重量weight了，所以从前到后遍历数组，找出所有的weight递增序列，序列总数即所耗时间time，具体看代码，已AC

#include<stdio.h>
#include<string.h>
typedef struct
{
    int l;
    int w;
} Stick;
Stick stick[5000];
bool flag[5000];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,i,j;
        memset(flag,false,sizeof(bool)*5000);
        scanf("%d",&n);
        for(i=0;i<n;i++)
            scanf("%d%d",&stick[i].l,&stick[i].w);
        for(i=0;i<n;i++)
            for(j=i+1;j<n;j++)
                if(stick[i].l>stick[j].l||(stick[i].l==stick[j].l&&stick[i].w>stick[j].w))
                {
                    Stick temp=stick[i];
                    stick[i]=stick[j];
                    stick[j]=temp;
                }
        int time=0;
        for(i=0;i<n;i++)//找序列 ,由于上面的已经将长度排好序列，现在只要比较宽度就OK了 ，就能找出一个满足条件的子序列 
        {
           if(flag[i]) continue;//stick[i]已经用过了 
           int w=stick[i].w;//子序列开头  第一个元素的重量 
           for(j=i+1;j<n;j++)//分别与其他元素比较 
           {
                  if(!flag[j]&&w<=stick[j].w)//符合条件，标记一下，防止找别的子序列时重复 
                  {
                         flag[j]=true;
                         w=stick[j].w;
                  }
           }
           time++;//找到一个只要1分钟序列， 
        } 
        printf("%d
",time);
    }
    return 0;
}

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