【HDOJ1051】【排序+LIS】【贪心】

http://acm.hdu.edu.cn/showproblem.php?pid=1051

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24757    Accepted Submission(s): 9973

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

题目大意：

有一堆n个木棍，长度质量已知，机器处理木棍需要设置时间，规定

（1）第一根木棍的设置时间是1min

（2）前一个处理的木棍长度和质量小于等于后一个就不用设置时间，否则需要1min设置

找到最小建立时间。

如 给出(4,9)(5,2)(2,1)(3,5)(1,4)则最小建立时间(1,4)(3,5)(4,9)(2,1)(5,2)。

（本来想找个贪心的水题耍一下..结果发现完全看不出这道题的贪心解法...然后想到之前做的一个导弹拦截的问题..感觉二者有异曲同工之妙..

题目分析：首先按照长度进行排序..长度相同的按照重量排序..然后就能直接取了？？（假的贪心..很容易找到hack数据..我的做法是在排序之后..从后往前找到这段数字的最长上升子序列的长度K,..则.K就是答案..这一点和导弹拦截的那题几乎一致。

导弹拦截http://acm.hdu.edu.cn/showproblem.php?pid=1257

 #include <bits/stdc++.h>

 using namespace std;

 struct pot{
     int len;
     int weig;
 }qwq[];
 bool cmp(struct pot aa,struct pot bb){
     if(aa.len!=bb.len)
         return aa.len < bb.len;
     return aa.weig < bb.weig;
 }
 int main()
 {
     int nums[];
     int t;
     scanf("%d",&t);
     while(t--){
         int n;
         scanf("%d",&n);
         memset(nums,,sizeof(nums));
        for(int i =  ; i < n; i++){
             scanf("%d%d",&qwq[i].len,&qwq[i].weig);
         }
         sort(qwq,qwq+n,cmp);
         int dp[];
         int tot=;
         dp[]=qwq[n-].weig;
         int maxx=qwq[n-].weig;
        for(int i = n- ; i >=  ; i--){
             if(qwq[i].weig>dp[tot]){
                 dp[++tot]=qwq[i].weig;
                 maxx=dp[tot];
             }
             else{
                 int l=;
                 int r=tot;
                 while(l<=r){
                     int mid=(l+r)/;
                     if(dp[mid]>=qwq[i].weig){
                         r=mid-;
                     }
                     else {
                         l=mid+;
                     }

                 }
                dp[l]=qwq[i].weig;
             }
        }
         cout << tot <<endl;
     }
     return ;
 }

（貌似导弹拦截那题大多数人也是使用贪心..（> _ <）