Wooden Sticks

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6563    Accepted Submission(s): 2707

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

 

 

此题排序之后，问题就可以简化，（假设按照长度不等时长度排，长度等是按照重量排，我假设按照从大到小来排！）即求排序后的所有的重量值最少能表示成几个集合。长度就不用再管了，从数组第一个数开始遍历，只要重量值满足条件，那么这两个木棍就满足条件！

还是需要注意一些问题的，比如第一次的代码里面标记就弄的很麻烦用single来标记，结果过不了。在第二次里面就用flag[]的数组来标记简化了，然而发现还是过不了。找了半天结果在输入想 1 5 1 6  和1 6 1 5的结果不一样，前一个是1，后一个是2.这才发现是快排的cmp()函数写错了，忽略了一种情况（当长度相同时，体重也要比较）。

第一次的麻烦代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define SIZE 5005
 struct Sticks
{
    int length;
    int weight;
};
 Sticks  stick[SIZE];
 int cmp(const void *a,const void *b)
 {
      return (*(Sticks *)a).length>(*(Sticks *)b).length ? 1:-1;
 }
int main()
{
    int T,n,i,j;
    int time;
    int max,single;
    scanf("%d",&T);
    while(T--){
        memset(stick,0,sizeof(struct Sticks)*SIZE);
        time=0;
        scanf("%d",&n);
        for(i=1;i<=n;i++){
            scanf("%d %d",&stick[i].length,&stick[i].weight);
        }
        qsort(stick+1,n,sizeof(stick[0]),cmp);
        for(i=1;i<=n;i++)
        {
            max=stick[i].weight;
            single=0;
            for(j=i;j<=n;j++)
            {
                if(max<=stick[j].weight&&max!=0){
                    max=stick[j].weight;
                    stick[j].weight=0;
                    single=1;
                }
                if(max==stick[i].weight&&max!=0){single=1;}
            }
            if(single==1){time++;}
        }
        printf("%d
",time);
    }
}

第二次改进后的代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define SIZE 5005
 struct Sticks
{
    int length;
    int weight;
};
 Sticks  stick[SIZE];
 int cmp(const void *a,const void *b)
 {
      return (*(Sticks *)a).length>(*(Sticks *)b).length ? 1:-1;
 }
int main()
{
    int T,n,i,j;
    int time;
    int max;
    int flag[SIZE];
    scanf("%d",&T);
    while(T--){
        memset(stick,0,sizeof(struct Sticks)*SIZE);
        time=0;
        scanf("%d",&n);
        for(i=1;i<=n;i++){
            scanf("%d %d",&stick[i].length,&stick[i].weight);
            flag[i]=0;
        }
        qsort(stick+1,n,sizeof(stick[0]),cmp);
        for(i=1;i<=n;i++)
        {
            if(flag[i]==1)  continue;
            max=stick[i].weight;
            for(j=i+1;j<=n;j++)
            {
                if(max<=stick[j].weight&&flag[j]==0){
                    max=stick[j].weight;
                    flag[j]=1;
                }

            }
            time++;
        }
        printf("%d
",time);
    }
    return 0;
}

第三次的正确代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define SIZE 5005
 struct Sticks
{
    int length;
    int weight;
};
 Sticks  stick[SIZE];
 /*int cmp(const void *a,const void *b)
 {
      return (*(Sticks *)a).length>(*(Sticks *)b).length ? 1:-1;
 }*/
 int cmp(const void *a,const void *b){
    Sticks c=*(Sticks *)a;
    Sticks d=*(Sticks *)b;
    if(c.length==d.length)
        return c.weight-d.weight;
    return c.length-d.length;
}
int main()
{
    int T,n,i,j;
    int time;
    int max;
    int flag[SIZE];
    scanf("%d",&T);
    while(T--){
        memset(stick,0,sizeof(struct Sticks)*SIZE);
        time=0;
        scanf("%d",&n);
        for(i=1;i<=n;i++){
            scanf("%d %d",&stick[i].length,&stick[i].weight);
            flag[i]=0;
        }
        qsort(stick+1,n,sizeof(stick[0]),cmp);
        for(i=1;i<=n;i++)
        {
            if(flag[i]==1)  continue;
            max=stick[i].weight;
            for(j=i+1;j<=n;j++)
            {
                if(max<=stick[j].weight&&flag[j]==0){
                    max=stick[j].weight;
                    flag[j]=1;
                }

            }
            time++;
        }
        printf("%d
",time);
    }
    return 0;
}