hdoj 1051 Wooden Sticks 【LIS变形】

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12976    Accepted Submission(s): 5344

Problem 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

 动规贪心都可以做。附上两种代码：

 

dp代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
struct wooden
{
    int l;
    int w;
}num[5010];
bool cmp(wooden a,wooden b)
{
    if(a.l!=b.l)
    return (a.l<b.l);
    else
    return (a.w<b.w);
}
int main()
{
    int t,n,i,j;
    int max;
    int dp[100000];
    scanf("%d",&t);
    while(t--)
    {
        max=0;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d%d",&num[i].l,&num[i].w);
            dp[i]=1;
        }
        sort(num,num+n,cmp);
        for(i=0;i<n;i++)
        {
            for(j=0;j<i;j++)
            {
                if(num[i].w<num[j].w&&dp[i]<dp[j]+1)
                dp[i]=dp[j]+1;
            }
        }
        for(i=0,max=dp[i];i<n;i++)
        {
            if(dp[i]>max)
            max=dp[i];
        }
        printf("%d\n",max);
    }
    return 0;
}

贪心代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
struct wooden
{
    int l;
    int w;
}num[5010];
bool cmp(wooden a,wooden b)
{
    if(a.l!=b.l)
    return (a.l<b.l);
    else
    return (a.w<b.w);
}
int main()
{
    int t,n;
    int k,i,j;
    int exist;
    int sequ[100000];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d%d",&num[i].l,&num[i].w);
            sequ[i]=1;
        }
        sort(num,num+n,cmp);
        k=1;sequ[0]=num[0].w;
        for(i=1;i<n;i++)
        {
            exist=1;
            for(j=0;j<k;j++)
            {
                if(sequ[j]<=num[i].w)
                {
                    sequ[j]=num[i].w;
                    exist=0;
                    break;
                }
            }
            if(exist)//前面所有重量大于当前重量，再开一个序列 
            {
                sequ[k++]=num[i].w;
            }
        }
        printf("%d\n",k);
    }
    return 0;
}