C - Wooden Sticks

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

 

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

这一题是学姐给的思路。就是先按照长度排序，然后再用flag记录是否访问过。

#include
         
          #include
          
           using namespace std;typedef struct dot{    int l,w;    bool flag;}dot;bool cmp(dot a,dot b){    if(a.l!=a.l)return a.w
           
            >T;    while(T--)    {        cin>>n;        dot a[10000];        for(int i=0;i!=n;i++){            cin>>a[i].l>>a[i].w;            a[i].flag=0;        }        sort(a,a+n,cmp);        int sum=0;        for(int i=0;i!=n;i++){            if(a[i].flag==0){                sum++;                a[i].flag=1;                int k=i;                for(int j=k;j!=n;j++){                    if(!a[j].flag&&a[k].w<=a[j].w){k=j;a[j].flag=1;}                }            }        }        cout<
            
             <
            
           
          
         

贪心练得不好。。。。