HDU 1051 Wooden Sticks

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

213

//Wooden Sticks 贪心#include <iostream>#include<algorithm>using namespace std;typedef struct stick{    int length;    int weight;    int flag;}St;St s[5003];int cmp(const St &a,const St &b){    if(a.length!=b.length) return a.length<b.length;    else a.weight<b.weight;}int main(){    int t;    int n;    cin>>t;    while(t--)    {        cin>>n;        for(int i=0;i<n;i++)        {            cin>>s[i].length>>s[i].weight;            s[i].flag=0;        }        sort(s,s+n,cmp);        int count=0;        for(int i=0;i<n;i++)        {            if(s[i].flag==1) continue;            count++;int tmp=i;            for(int j=i+1;j<n;j++)            {                if(!s[j].flag&&s[j].length>=s[tmp].length&&s[j].weight>=s[tmp].weight)                {                    s[j].flag=1;                    tmp=j;                }            }        }        cout<<count<<endl;    }  return 0;}