HDU1051：Wooden Sticks

Problem DescriptionThere 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). 

InputThe 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. 

OutputThe 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;struct Node{    int head;    int end;    int flag;} node[5005];int cmp(struct Node a,struct Node b){    if(a.head < b.head)    return 1;    else if(a.head == b.head)    return a.end < b.end;    return 0;}int main(){    int n,t;    cin >> t;    while(t--)    {        cin >> n;        int i,j,sum = 0,cnt = 0;        for(i = 0; i<n; i++)        {            cin >> node[i].head >> node[i].end;            node[i].flag = 0;        }        sort(node,node+n,cmp);        int ans = 0;        for(i = 0; i<n; i++)        {            if(!node[i].flag)            {                node[i].flag = 1;                ans++;                int end = node[i].end;                for(int j = i+1; j < n; j++)                {                    if(!node[j].flag && node[j].end >= end)                    {                        node[j].flag = 1;                        end = node[j].end;                    }                }            }        }        cout << ans << endl;    }    return 0;}