【杭电1051】升序子序列个数

 Wooden Sticks

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status

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
有一堆棍子，棍子有长度（L）和重量（W）。机器处理第一根棍子花费1秒。如果后一根棍子的长度（L`)和重量（W`)都比前一根的大，则后一根棍子不需要花费时间（L`>=L&&W`>=W)。
求怎样处理这一堆棍子，所花费的时间最少。
本题可以运用贪心策略，首先按照L进行递增排序，如果L相等则按照W递增进行排序（当然也可以把W当做第一关键字，L当做第二关键字）。
排序后，从前往后，找出每一根棍子的后继棍子（条件为后一根棍子的L和W都>=前一根的）。

     

      

       
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#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
	int l,w;
}x[5001];
bool cmp(node a,node b)
{
	if(a.l!=b.l)
	return a.l<b.l;
	else
	return a.w<=b.w;
}
int vis[5001];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		scanf("%d",&n);
		for(int i=0;i<n;i++)
		{scanf("%d%d",&x[i].l,&x[i].w);
		vis[i]=0;
		}
		sort(x,x+n,cmp);
		int count=0;
		for(int i=0;i<n;i++)
		{
		if(vis[i]==1)//已经遍历过，跳过
				continue;
				count++; 
				int q=i;
			for(int j=i+1;j<n;j++)
			{
				
				if(vis[j]==0&&x[j].w>=x[q].w)
				{
				vis[j]=1; q=j;//遍历一组符合要求不需要花费时间的点 
				}
			}
		}
		printf("%d\n",count);
	}
	return 0;
}
       

        还是没想通我写的哪里出问题了o(︶︿︶)o
       
       

        

       
       

        
        
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
	int a,b;
}x[5001];
bool cmp(node x,node y)
{
	if(x.a!=y.a)
	return x.a<y.a;
	else
	return x.b<=y.b;
}
int main()
{
	int t,n;
	scanf("%d",&t);
	while(t--)
	{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d%d",&x[i].a,&x[i].b);
		sort(x+1,x+n+1,cmp);
		int f=1,t=1;
		for(int i=2;i<=n;i++)
		{
			for(int j=1;j<i;j++)
			{
				if(x[j].a<x[i].a&&x[j].b<x[i].b)//前面有比他小的，归为一组，不需要花费时间 
				{
				f=0;break;	
				}	
			}
			if(f==1)
				++t;
		}
		printf("%d\n",t);
	 } 
	return 0;
}