Wooden Sticks

Wooden Sticks

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

Submit Status

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 

 思路：这道题。。困扰我好久。看了网上的思路和同学的想法，终于AC了。。

 首先，想让扣的时间最少，所以必须想到，长度和重量最好都要先保证比较小，然后逐渐做质量和长度稍大的。

先对长度进行升序，如果长度相同的话，就对重量进行升序。（长度可能会相同，所以比较时候要注意，我当时没有想到，错了2次）。例如 1 4 1 3 应该输出1，但是如果只看长度的话，可能会输出2。

长度非递减排列的情况下，判断重量，如果重量也是前者小于等于紧挨的后者，就标记为true（开始这么多产品都标记为false），如果不满足不处理。。进行了一次总体的遍历。然后再来“一遍”，从左往右，使被标记为false 的产品进行比较，开始要加一个时间，最左边的false那个产品标记为true，如果前者的重量“紧挨着”小于等于后者，后者就变成true，否则不处理。继续进行循环。。。

#include<iostream>#include<string.h>#include<algorithm>using namespace std;struct stick{int length;int weight;};bool cmp(stick a,stick b){if(a.length!=b.length)return a.length<b.length;else  return a.weight<b.weight;}int main(){int T,n,i,j,k;stick a[10000];bool b[10000];cin>>T;while(T--){int sum=1;memset(b,false,sizeof(b));cin>>n;for(i=0;i<n;i++){cin>>a[i].length>>a[i].weight;}    sort(a,a+n,cmp);int count=0;      for(i=0;i<n;i++)  {   if(b[i]==false)  {  count++;  b[i]=true;    j=i;  for(k=j+1;k<n;k++)  {  if(b[k]==false)  {  if(a[j].weight<=a[k].weight)  {  b[k]=true;  j=k;  }  }  }  }  }  cout<<count<<endl;}return 0;}