Wooden Sticks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11023 Accepted Submission(s): 4530
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
【题目来源】
【题目大意】
有一堆木棒,这些木棒的长度和重量是给定的。这些木棒将被机器一根接一根的进行处理。机器处理每一根木棒需要一些时间,这个时间被叫做建立时间。这个时间是这样定义的:1.第一根木棒的建立时间是1分钟;2.在处理一根长度为l、重量为w的木棒之后,如果木棒的l'<=l、w<=w',那么这个机器将不需要建立时间。否则,这个机器将需要1分钟的建立时间。你需要找到处理完这些木头的最短建立时间。题目中的例子看懂就知道怎么做了。【题目分析】
首先按照l排序,当l相同的时候,再按照w排序,然后就是按照顺序一遍一遍的标记并统计,直至所有的都被标记,最后的count就是答案。
奇怪的是,我的代码在测试样例的时候是错的,我提交后竟然ac了。
#include#include #include using namespace std;struct Node{ int l,w; bool mark;};bool cmp(Node a,Node b){ if(a.l==b.l) { return a.w =cl&&node[j].w>=cw) { node[j].mark=1; cl=node[j].l; cw=node[j].w; } } j=1; while(node[j].mark) j++; i=j; if(i==n) break; cnt++; node[i].mark=1; cl=node[i].l; cw=node[i].w; } printf("%d\n",cnt); } return 0;}