NO.1065: Wooden Sticks
Time Limit:1000MS Memory Limit:10000K
Total Submit:2716 Accepted:1185

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 ( 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

 
 
#include <iostream>
using namespace std;

class Node{
public:
 int length;
 int weight;
 bool flag;
};

void main()
{
 int T;
 cin>>T;
 if(T<=0) cout<<0<<endl;
 int assist=T;
 while(assist)
 {
  int n;
  cin>>n;
  if(n<=0)
  {
   cout<<0<<endl;
   continue;
  }

  Node *node=new Node[n];
  int *iter=new int[n];

  int i,j;
  for(i=0;i<n;i++)
  {
   cin>>node[i].length>>node[i].weight;
   iter[i]=i;
   node[i].flag=1;
  }

  for(i=0;i<n;i++)
  {
   for(j=i+1;j<n;j++)
   { 
    int temp;
    if(node[iter[i]].length > node[iter[j]].length)
    {
     temp=iter[i];
     iter[i]=iter[j];
     iter[j]=temp;
    }
    else if(node[iter[i]].length == node[iter[j]].length)
    { 
     if(node[iter[i]].weight > node[iter[j]].weight)
     {
      temp=iter[i];
      iter[i]=iter[j];
      iter[j]=temp;
     }
    }
   }
  }

  int count=0;
  for(i=0;i<n;i++)
  {
   if(node[iter[i]].flag==1)
   {
    count++;
    node[iter[i]].flag=0;
   }

   for(j=i+1;j<n;j++)
   {
    if( node[iter[i]].weight < node[iter[j]].weight )
    {
     node[iter[j]].flag=0;
    }
   }
   
  }

  cout<<count<<endl;
  delete []node;
  delete []iter;
  assist--;
 }
}
 
我的为什么就是通不过它的在线评测啊？我搞了一下午也没想出问题的所在，有没有高人指点一下

[此贴子已经被作者于2007-7-15 17:19:36编辑过]

what is your idea? Don't want to read code.

简单点说，就是一贪心策略， sticks 按length升序排列，而后，两个for循环，逐个考查是否需要建立时间。（flag字段表示是否仍需要考查）

这题就是贪心啊，你怎么写那么多

#include <iostream>
#include <list>
#include <algorithm>
using namespace std;

struct Node{
    int l;
    int w;
};    
bool operator < (const Node& a,const Node& b)
{
    if(a.l!=b.l)return (a.l <b.l);
    return (a.w<b.w);
}
int main()
{
    int N,n;
    scanf(\"%d\",&N);
    while(N--){
        scanf(\"%d\",&n);
        list<Node> L;
        Node tempnode;
        for(int i=0;i<n;i++){
            scanf(\"%d%d\",&tempnode.l,&tempnode.w);
            L.push_back(tempnode);
        }
        L.sort();
        
        int cnt=0,pre_w;
        while(!L.empty()){
            list<Node>::iterator p;
            ++cnt;
            for(p=L.begin(),pre_w=p->w;p!=L.end();){
                if(p->w>=pre_w){
                    pre_w=p->w;
                    p=L.erase(p);
                }
                else {
                    ++p;
                }    
            }
        }
        printf(\"%d\n\",cnt);
    }    
    return 0;
}

#include <iostream>
using namespace std;

class Node{
public:
int length;
int weight;
bool flag;
};

void main()
{
int T;
cin>>T;

while(T--)
{
int n;
cin>>n;

Node *node=new Node[n];

int i,j;
for(i=0;i<n;i++)
{
cin>>node[i].length>>node[i].weight;
node[i].flag=1;
}


//按length排序：
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
int temp1,temp2;
if(node[i].length>node[j].length)
{
temp1=node[i].length; temp2=node[i].weight;
node[i].length=node[j].length; node[i].weight=node[j].weight;
node[j].length=temp1; node[j].weight=temp2;
}
else if(node[i].length==node[j].length)
{
if(node[i].weight>node[j].weight)
{
temp1=node[i].length; temp2=node[i].weight;
node[i].length=node[j].length; node[i].weight=node[j].weight;
node[j].length=temp1; node[j].weight=temp2;
}
}
}
}

int count=0;
for(i=0;i<n;i++)
{
if(node[i].flag==1)
{
count++;
node[i].flag=0;
}

for(j=i+1;j<n;j++)
{
if( node[i].weight <= node[j].weight )
{
node[j].flag=0;
}
}
}

cout<<count<<endl;
delete []node;
}
}