hdu 1051 Wooden Sticks

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23703    Accepted Submission(s): 9625

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

 

 

类似最少拦截系统，求最长不降子序列

 

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
struct node
{
    int l, w;
}a[5005];
bool cmp(node x, node y)
{
    if (x.l == y.l)
        return x.w < y.w;
    return x.l < y.l;
}
int main()
{
    int dp[5005];
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        int i;
        for (i = 1; i <= n; i++)
        {
            cin >> a[i].l >> a[i].w;
            dp[i] = 1;
        }
        sort(a + 1, a + 1 + n, cmp);
        int j;
        int k = 1;
        int ans =0;
        for (i = 2; i <= n; i++)
        {
            for (j =1; j<=i; j++)
            {
                if (a[j].w>a[i].w)//前面有一个比它要大
                {
                    dp[i] =max(dp[i], dp[j]+1);
                }
            }
            ans = ans > dp[i] ? ans : dp[i];
        }
        cout << ans << endl;
    }
    return 0;
}