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

213

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(constvoid *a,constvoid *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;

}

intmain()

{

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);

}

return0;

}