Wooden Sticks

Wooden Sticks

Crawling in process...Crawling failedTime Limit:1000MS    Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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

213 
#include <iostream>#include<algorithm>using namespace std;#include <string.h>struct stick///这个结构体存储这个棍子的所有信息{    int l;    int w;} s[5000];bool cmp(stick a,stick b)///这个是按照棍子的长度来排序，如果长度一样那么就按照轻的在前面的顺序来排序{    if(a.l<b.l)        return true;    else if(a.l>b.l)        return false;    else///这是如果相等的情况下        return a.w<b.w;}int main(){    int t,n,minute,number,i;    int mark[5000];///标记数组，标记这根木棒有没有被加工过    cin>>t;    while(t--)    {        cin>>n;        for( i=0; i<n; i++)            cin>>s[i].l>>s[i].w;        sort(s,s+n,cmp);        memset(mark,0,sizeof(mark));        number=0;        minute=0;        int pl;        while(number!=n)        {            for( i=0; i<n; i++) ///这个循环是找个好的开头，从小到达找                if(!mark[i])                {                    pl=i;                    minute++;                    break;///找到一个就出去，找到一个最短的                }            for( i=0; i<n; i++) ///遍历寻找下一个满足条件的，也就是不要时间的，一次性找完            {                if(!mark[i]&&s[i].l>=s[pl].l&&s[i].w>=s[pl].w)///这个就是要满足的条件                {                    mark[i]=1;                    number++;                    pl=i;                }            }        }        cout<<minute<<endl;    }    return 0;}