Wooden Sticks

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

这个题意很简单，容易看懂，就是看有多少个连续递增的子序列。

DP:

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[5009];
struct node{
int l,w;
}s[5009];
int cmp(node a,node b)
{
if(a.l !=b.l )
return a.l <b.l ;
return a.w <b.w ;
}
int main()
{
int t,n,i,j;
scanf("%d",&t);
while(t--)
{
int ans=0;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d%d",&s[i].l ,&s[i].w );
dp[i]=1;
}
sort(s,s+n,cmp);
for(i=0;i<n;i++)
{
for(j=0;j<i;j++)
{
if(s[i].w <s[j].w &&dp[i]<dp[j]+1)
dp[i]=dp[j]+1;
}
ans=max(ans,dp[i]);
}
printf("%d\n",ans);
}
return 0;
}

贪心：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[5009];
struct node{
int l,w;
}s[5009];
int cmp(node a,node b)
{
if(a.l !=b.l )
return a.l <b.l ;
return a.w <b.w ;
}
int main()
{
int t,n,i,j,flag;
scanf("%d",&t);
while(t--)
{
memset(dp,0,sizeof(dp));
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d%d",&s[i].l ,&s[i].w );
sort(s,s+n,cmp);
int xulie=1;dp[0]=s[0].w ;
for(i=1;i<n;i++)
{
flag=1;
for(j=0;j<xulie;j++)
{
if(s[i].w >=dp[j])
{
flag=0;
dp[j]=s[i].w ;
break;
}
}
if(flag)
dp[xulie++]=s[i].w ;
}
printf("%d\n",xulie);
}
return 0;
}