Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11249    Accepted Submission(s): 4629

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

 

Source

Asia 2001, Taejon (South Korea)

 

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在POJ呆惯了，WA一直以为是代码问题。没想到是数组开小了。。。。
这道题关键在于为什么贪心能够找到最优解。
刚開始我不懂，为什么用贪心能够找出最优解！也在这个问题上纠结了非常久，感觉比較痛苦！
这和找最长递增子序列不同，仅仅要能增加到原有的序列中比新开一个递增子序列，所得到的子序列数一定更少。

#include <iostream>
#include <algorithm>
using namespace std;
#define M 11000
struct node{
    int l,w;
}f[M];
int vis[M];
//将长度排序，减少成为一位数组的扫描。 
bool cmp(node a,node b){
    if(a.l<b.l) return true;
    if(a.l>b.l) return false;
    if(a.l==b.l) return a.w<b.w;
    
}
int main()
{
    int n,m,t,i,j,cur,tot;
    while(scanf("%d",&n)!=EOF)
    {   
        while(n--)
        {   
            cur=0;tot=0;
            memset(vis,0,sizeof(vis));
            scanf("%d",&m);
            for(i=0;i<m;i++)
            {  
                scanf("%d%d",&f[i].l,&f[i].w);
            }
            sort(f,f+m,cmp);
            for(i=0;i<m;i++)
            {  
                if(vis[i]) continue; //假设已经排入一个递增子序列，就不用再考虑。

                t=f[i].w;vis[i]=1;
                for(j=i+1;j<m;j++)
                {   
                    if(!vis[j]&&t<=f[j].w)        //找递增子序列的元素。
                    {
                        vis[j]=1;
                        t=f[j].w;
                    }    
                }
                tot++;
            }
            printf("%d\n",tot);
        }
    } 
    return 0;
}