贪心 木棍问题

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10140    Accepted Submission(s): 4169

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

213

 

#include<stdio.h>  #include<string.h>  #include<algorithm>  using namespace std;  struct sa  {      int l;      int w;  }data[10000];  int cmp(sa a,sa b)  {      if(a.l!=b.l)      return a.l<b.l;      return a.w<b.w;  }  int z[10000];  int main()  {      int n;      while(scanf("%d",&n)!=EOF)      {          for(int a=0;a<n;a++)          {              int t;              scanf("%d",&t);              for(int b=0;b<t;b++)              {               scanf("%d%d",&data[b].l,&data[b].w);              }              sort(data,data+t,cmp);              memset(z,0,sizeof(z));              int x=0,y=0,i=0,j=0,time=0,k=0;              while(k<t)              {                 for(i=j;i<t;i++)                 if(z[i]==0)                 {                     z[i]=1;                     j=i+1;                     x=data[i].l;                     y=data[i].w;                     k++;                     break;                 }                 for(i=j;i<t;i++)                 {                     if(z[i]==0&&x<=data[i].l&&y<=data[i].w)                     {                        z[i]=1;                        x=data[i].l;                        y=data[i].w;                        k++;                     }                   }                 time++;              }              printf("%d\n",time);          }        }  }