Wooden Sticks

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 ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .

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 2n 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.

The output should contain the minimum setup time in minutes, one per line.

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 

213

Taejon 2001

 

 

#include<iostream>#include<algorithm>#include<string.h>using namespace std;struct Node{    int x;    int y;};int cmp(Node a,Node b){    if(a.x==b.x) return a.y<b.y;    return a.x<b.x;}int main(){    int t;    int n;    Node node[5005];    bool flag[5005]; //标志是否已经分组    while(cin>>t)    {        while(t--)        {            memset(flag,0,sizeof(flag));            cin>>n;            int ans=0;            for(int i=0;i<n;i++)                cin>>node[i].x>>node[i].y;            sort(node,node+n,cmp);            for(int i=0;i<n;i++)            {                if(flag[i]==0)                {                    int p=node[i].y;                    for(int j=i;j<n;j++)                    {                        if(flag[j]==0&&node[j].y>=p) {flag[j]=1;p=node[j].y;}                    }                    ans++;                }            }            cout<<ans<<endl;        }    }}