杭电ACM1008

(1)题目大意:
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 ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .
 

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 2n 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
题意是:将木棍放在机器里处理,第一根需要一分钟,剩余的如果大于等于前边放入的长度和重量,就不用费时间,否则需要一分钟,计算给出一组数的最少时间!用贪心算出最少降序序列个数!
(2)思路:
  首先根据长度进行排序,然后在对重量进行比较,如果下一根小,则时间不加一,否则时间加以而且将此重量赋值。
(3)感想:
  我就不明白了,这个题跟1001一模一样,我自己写的1001提交就对了,看着题目完全一样,就用以前的代码放上了,给我提示编译错误!!compile error,还真是日了狗了,所以我用1001的题目名字搜了一份代码,交在1008上,结果通过了,这里说明两份提完全一样的!!真不明白,日了狗了!
(4)代码:
自己的,跟1001一样:
#include<iostream>
#include<algorithm>
using namespace std;
struct My
{
 int Wight, Long;
 bool f = false;
 bool operator<(const My &x)
 {
  if (x.Long == Long)
  {
   return Long < x.Long;
  }
  else
  {
   return Long < x.Long;
  }
 }
};
int main()
{
 int n, m, mm = 0;
 My my[5000];
 cin >> m;
 while (mm < m)
 {
  cin >> n;
  for (int i = 0;i < n;i++)
  {
   cin >> my[i].Wight;
   cin >> my[i].Long;
   my[i].f = false;
  }
  sort(my, my + n);
  int time = 0;
  for (int i = 0;i < n; i++)
  {
   if (my[i].f == true)
   {
    continue;
   }
   else
   {
    time++;
    my[i].f == true;
   }
   for (int j = i + 1;j < n;j++)
   {
    if (my[j].f == true)
    {
     continue;
    }
    else
    {
     if (my[j].Wight >= my[i].Wight&&my[j].Long >= my[i].Long)
     {
      my[j].f = true;
      my[i].Long = my[j].Long;
      my[i].Wight = my[j].Wight;
     }
     else
     {
      continue;
     }
    }
   }
  }
  cout << time << endl;
  mm++;
 }
}
网上找的:
#include<stdio.h>
#include<stdlib.h>
typedef struct stu {
int l ,w ;
}Lode ;
Lode s [ 5005 ];
int cmp (const void *a ,const void *b )
{
Lode *c ,*d ;
c =(Lode *)a ;
d =(Lode *)b ;
if(c ->l !=d ->l )
return d ->l -c ->l ;
else return d ->w -c ->w ;
}
int main()
{
int t ,n ,i ,j ,ans ;
scanf ( "%d" ,&t );
while(t --)
{
scanf ( "%d" ,&n );
for(i = 0 ;i <n ;i ++)
scanf ( "%d%d" ,&s [i ].l ,&s [i ].w );
qsort (s ,n ,sizeof(s [ 0 ]),cmp );
ans =n ; for(i = 1 ;i <n ;i ++)
for(j = 0 ;j <=i - 1 ;j ++)
{
if(s [j ].l >=s [i ].l &&s [j ].w >=s [i ].w )
{
ans --;
s [j ].l =s [i ].l ;
s [j ].w =s [i ].w ;
s [i ].l = 0 ;
s [i ].w = 0 ;
break;
}
}
printf ( "%d\n" ,ans );
}
return 0 ;
}

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