01065: Wooden Sticks

greedy, binary search http://cs101.openjudge.cn/practice/01065/

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 ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .

输入

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 2n 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.

输出

The output should contain the minimum setup time in minutes, one per line.

样例输入

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

样例输出

2
1
3

来源: Taejon 2001

python

def min_setup_time(sticks):
    n = len(sticks)
    check = [0]*n
    setup_time = 0
    while (0 in check):
        #print(check)
        #print(sticks)
        i = 0
        for j in range(n):
            if check[j] == 0:
                i = j
                break
        current = sticks[i]
        check[i] = 1
        setup_time += 1
        i += 1
        while  i < n:
            if  check[i]==0 and current[0]<=sticks[i][0] and  current[1]<= sticks[i][1]:
                check[i] = 1
                current = sticks[i]
            
            i +=1

    return setup_time


T = int(input())  
for _ in range(T):
    n = int(input())
    data = list(map(int, input().split()))
    sticks = [(data[i], data[i + 1]) for i in range(0, 2 * n, 2)]
    sticks.sort()
    print(min_setup_time(sticks))

python

#dilworth和最长单调子序列
# 答案就是对l排序后求w的最长严格递减子序列（用Dilworth's theorem不难证明），
# 最长严格递减子序列有经典的nlogn的算法
# （https://en.wikipedia.org/wiki/Longest_increasing_subsequence）。
#一般有一样的都不是大问题，因为可以把(3,5) (3,6) 直接看作(3.1, 5) (3.2, 6)

import bisect

def doit():
    n = int(input())
    data = list(map(int, input().split()))
    sticks = [(data[i], data[i + 1]) for i in range(0, 2 * n, 2)]
    sticks.sort()
    f = [sticks[i][1] for i in range(n)]
    f.reverse()
    stk = []

    for i in range(n):
        t = bisect.bisect_left(stk, f[i])
        if t == len(stk):
            stk.append(f[i])
        else:
            stk[t] = f[i]

    print(len(stk))

T = int(input())
for _ in range(T):
    doit()

【陈子良 25物理学院】这题让我想到了T25353排队，都是每次从列表中选出满足一定条件的最多的元素。首先将木棍按l排序加入队列中，每次遍历一遍队列元素，如果某一元素的w大于当前值，就能在此次加工，否则重新加入队列中。

python

from collections import deque
for _ in range(int(input())):
    n=int(input())
    iter1=iter(map(int,input().split()))
    sticks=[]
    for _ in range(n):
        sticks.append((next(iter1),next(iter1)))
    sticks.sort()
    queue=deque(sticks)
    s=0
    while queue:
        m=-float('inf')
        for _ in range(len(queue)):
            a,b=queue.popleft()
            if b>=m:
                m=b
            else:
                queue.append((a,b))
        s+=1
    print(s)

01065: Wooden Sticks ​

01065: Wooden Sticks