【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[5005];
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[405];
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        memset(nums,0,sizeof(nums));
       for(int i = 0 ; i < n; i++){
            scanf("%d%d",&qwq[i].len,&qwq[i].weig);
        }
        sort(qwq,qwq+n,cmp);
        int dp[5005];
        int tot=1;
        dp[1]=qwq[n-1].weig;
        int maxx=qwq[n-1].weig;
       for(int i = n-2 ; i >= 0 ; i--){
            if(qwq[i].weig>dp[tot]){
                dp[++tot]=qwq[i].weig;
                maxx=dp[tot];
            }
            else{
                int l=1;
                int r=tot;
                while(l<=r){
                    int mid=(l+r)/2;
                    if(dp[mid]>=qwq[i].weig){
                        r=mid-1;
                    }
                    else {
                        l=mid+1;
                    }

                }
               dp[l]=qwq[i].weig;
            }
       }
        cout << tot <<endl;
    }
    return 0;
}

（貌似导弹拦截那题大多数人也是使用贪心..（> _ <）