poj 1065 Wooden Sticks (dp)

问题给出n个木头， 每个木有有li ， wi的属性， 如果后一根木头的li>=li-1 且wi>=w-1 那么不需要等加工时间， 否则加1.

第一根木头总要1个时间的加工时间，问最少需要多少时间

首先根据木头的某一个属性排序， 然后求 最长下降子序列（ 等于求出最长不下降的 个数）

题目：

Wooden Sticks

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17044		Accepted: 7121

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

Source

Taejon 2001

 

 

代码：

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

int n;

struct W
{
        int l, w;
        bool operator < (const W &x )const
        {
            if( w< x.w)
                return true;
            else if ( w == x.w && l < x.l)
                return true;


            return false;
        }
};

W w[5000+10];
int dp[10000+10];
void init()
{
    cin>>n;

    for(int i=0;i<=n;i++)
    {
        dp[i] = 0;
    }
    for(int i=0;i<n;i++)
    {
        cin>>w[i].l>>w[i].w;
    }
    sort(w,w+n);
}
int main()
{

    int tst;
    cin>>tst;
    while(tst--)
    {
        init();
        int cnt = 0;
        for(int i=0;i<n;i++)
        {
            dp[i] =1;
            for(int j=0;j<i;j++)
            {
                if( w[j].l>w[i].l)
                {
                    dp[i] = max(dp[i],dp[j]+1);
                }
                cnt = max( cnt, dp[i]);
            }

        }
        cout<<cnt<<endl;
    }


    return 0;
}