HDU-1051 Wooden Sticks

一 题意描述：

                                                Wooden Sticks

　　　　　　　　　　　　　　　　　　      Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
　　　　　　　　　　　　　　　　　　　　　　　　          Total Submission(s): 3902    Accepted Submission(s): 1621

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

2
1
3

二 思路分析：

本题考察贪心算法的运用，我们可以把数据按照长度从小到大的顺序排序，如果长度相等那么就按照质量从小到大的顺序排序。我们在采用两个for循环进行搜索的时候建议用一个标志数组,如果已经被访问直接跳过。

三 代码展示：

#include <iostream>
#include <algorithm>
#include<iomanip>
#include<string.h>
using namespace std;
#define maxn 5005
struct stick
{
   int len,weight;
}st[maxn];
int judge[maxn];
int cmp(stick a,stick b)
{
    if(a.len!=b.len) return a.len<b.len;//如果长度不等按照长度排序
    else
    {
        return a.weight<b.weight;//如果长度相同按照质量大小排序
    }
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        int total=0;
        for(int i=0;i<n;i++)
        cin>>st[i].len>>st[i].weight;
        sort(st,st+n,cmp);//对结构体数组进行排序
        memset(judge,0,sizeof(judge));//标志数组置为0
        for(int i=0;i<n;i++)
        {
            if(judge[i]) continue;
            int min=st[i].weight;
               for(int j=i+1;j<n;j++)
              {
                  if(st[j].weight>=min&&judge[j]==0)//质量比起始值大并且未被访问过
                 {
                  judge[j]=1;min=st[j].weight;//往后搜寻注意及时更新min的值。
                 }
               }
               total++;
        }
        cout<<total<<endl;
    }
    return 0;
}