HDU1051Wooden Sticks

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15722    Accepted Submission(s): 6462

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：

 有一捆木棍，木棍的长度和重量是提前给定了的，现在要将这些木棍放入机器中去加工，加工需要花费一定的时间，问要求最小的加工时间，加工的时间规则如下：

（1）第一根木棍的加工时间是一分钟；

（2）接下来的木棍如果长度 重量和前一根木棍的关系为 L2>=L1并且W2>=W1，那么这根木棒加工的时间是不需要重新计算的，否则就要重新算一根木棍加工的时间；

省题：

由题目意思可以得到，我们要求把每次花费的加工时间是最少的，那么我们需要将满足第二条加工时间的规则的序列依次找出来，并找出能够满足题意的这种序列的个数，这就是我们所需要花费的时间；

思路：

对每根木棍进行长度的升序排序，如果木棍的长度相同，则比较其重量；然后将给每一根木棍标记为0；每次将没有比较过的进行比较，然后更新比较值；

如下：

对于题目给定数据：4 9 5 2 2 1 3 5 1 4

易知：

初始值： L : 4   5   2  3  1  

W: 9   2  1  5  4

排序后： L : 1 2 3 4 5

W: 4   1  5   9  2

标记值 flag：0 0 0 0 0

遍历第一次： L : 1   2   3  4  5 
W：4 1 5 9 2

flag： 1 0 1 1 0 count=1；

遍历第二次：

L：1 2 3 4 5

W: 4 1 5 9 2

flag: 1 1 1 1 1 count=2;

给出代码： #include<iostream> #include<cstring> #include<algorithm> #include<cstdio> using namespace std; void Swap(int &a, int &b)//交换函数 { int t = a; a = b; b = t; } int main() { int m, n; int L[5005], W[5005]; cin >> m; while (m--) { memset(L, 0, sizeof(L)); memset(W, 0, sizeof(W)); cin >> n; for (int i = 0; i < n; i++) { cin >> L[i] >> W[i]; } for (int i = 0; i < n - 1; i++)//对木棍进行长度排序，长度一样的比较重量； { for (int j = i + 1; j < n; j++) if (L[i]>L[j]) { Swap(L[i], L[j]); Swap(W[i], W[j]); } else if (L[i] == L[j] && W[i]>W[j]) { Swap(L[i], L[j]); Swap(W[i], W[j]); } } int flag[5005]; memset(flag, 0, sizeof(flag)); int count = 0; for (int i = 0; i < n; i++) //遍历得到最少需要花费的时间； { if (flag[i] == 0) { count++; int s = L[i]; int k = W[i]; for (int j = i + 1; j < n; j++) { if (flag[j] == 0 && s <= L[j] && k <= W[j]) { s = L[j]; k = W[j]; flag[j] = 1; } } } } cout << count << endl; } return 0; } </cstdio></algorithm></cstring></iostream>