zoj 1025 Wooden Sticks

Wooden Sticks

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

Output for the Sample Input

2 
1 
3

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先按长度排序，然后再计算重量的最长单调递增序列的个数。

#include <iostream>
#include <cstdio>
#include <string.h>
#include <algorithm>
using namespace std;
struct node
{
    int L;
    int w;
    bool flag;
};
int cmp(const void *a,const void *b)
{
    node *a1=(node *)a;
    node *b1=(node *)b;
    if(a1->L !=b1->L)
    {
        return a1->L-b1->L;
    }
    return a1->w-b1->w;
}
int main()
{
    int Case,ans;
    node a[5001];
    int b[5001];
    int n;
    cin>>Case;
    while (Case--)
    {
        cin>>n;
        ans=1;
        for (int i=0; i<n; i++)
        {
            cin>>a[i].L>>a[i].w;
            a[i].flag=true;
        }
        qsort(a, n, sizeof(node), cmp);
       
        for (int i=0; i<n; i++)
        {
            a[i].flag=false;
            b[i]=1;
            for (int j=i; j<n-1; j++)
            {
              if(a[j].w>=a[i].w  && a[j].flag  && a[j].w<=a[j+1].w)
              {
                  a[j].flag=false;
                  b[i]++;
              }
            }
        }
       
        memset(b,0, sizeof(b));
        b[0]=1;
        int k;
        for (int i=1; i<n; i++)
        {
            k=0;
            for (int j=0; j<i; j++)
            {
                if(a[j].w<a[i].w  && k<b[j])
                {
                    k=b[j];
                }
            }
            b[i]=k+1;
        }
        int Max=-1;
        for (int i=0; i<n; i++)
        {
            if(b[i]>Max)
            {
                Max=b[i];
            }
        }
        cout<<Max<<endl;
    }
    return 0;
}