2016SDAU课程练习一1001 Problem B

Problem B

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 147   Accepted Submission(s) : 42

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

 大意就是切木头，第一个要1min准备时间，下一根如果l和w任意一个小于前一根就要重新准备，求最短时间。

典型的贪心，现在感觉这种传统的贪心题做起来越来越得心应手了，感觉贪心不难，难的是1、对数据排序2、处理数据，这道题本来想的是直接全排好序，然后直接判断 ，但是发现难以实现，因为两组数据无法同时排序，只能优先一组，于是选择排一组，算一组。

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<set>
#include<algorithm>
using namespace std;
struct wood{
    int l;
    int w;
}woods[5010];
int num[5050];
bool cmp(const wood &a,const wood &b){
    if(a.l!=b.l)
        return a.l>b.l;
    else
        return a.w>b.w;
}
int main(){
    int m,n,min;
    cin>>m;
    for(int i=0;i<m;i++){
        cin>>n;
        for(int k=0;k<n;k++){
            cin>>woods[k].l>>woods[k].w;
            num[k]=0;
        }
        sort(woods,woods+n,cmp);
        int sum=0;
  for(int k=0;k<n;k++){
   if(num[k])
                continue;
   min=woods[k].w;
   for(int j=k+1;j<n;j++)
   {
    if(min>=woods[j].w && !num[j])
    {
     min=woods[j].w;
     num[j]=1;
    }
   }
            sum++;
        }
        cout<<sum<<endl;
    }
    return 0;
}