Wooden Sticks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14161 Accepted Submission(s): 5860
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
Source
Asia 2001, Taejon (South Korea)
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今天是端午节,祝各位C友端午节安康!好!废话不多说,开始我们的每日一水时间!
终于来了一道不是那么水的题了,首先解释一下题目的大意:
题目大意:
给你n根木棍的长度和重量。根据要求求出制作该木棍的最短时间。建立第一个木棍需要1分钟,如果接着制作的木棍比这个木棍的长度长,重量要重,那么接着制作的木棍不需要花费时间!然后如果再继续接着制作,则下一个木棍要比上一个木棍的长度长,重量大,则这个木棍也不需要花费时间!依次类推,反之,则需要花费一分钟,然后让你求出制作这一批木棍花费的最少的时间是多少!
解题思路:
其实大概明白了题意,便可以知道,这题考的是贪心!然后关键就在于贪心策略的选择!题目说了求最小时间,那就肯定要求长度和重量都要相对来说最小,我的策略如下:
1.首先按照木棍的长度进行排序,如果长度相等则重量重的放在前面。
2.当然排序完后,不一定都是下一根木棍重量和长度都大于前一根的,但是我们可以根据排序建立的数组再进行多次扫描,比较。
3.根据建立的数组依次比较他们的重量,如果大于标记数组跳过,否则ans加1。
#include<cstdio> #include<iostream> #include<algorithm> using namespace std; const int maxn = 5010; struct stick { int length; int weight; bool vis; }; bool cmp(stick s1,stick s2) { if(s1.length==s2.length) return s1.weight<s2.weight; else return s1.length<s2.length; } int n; stick s[maxn]; int main() { int t; scanf("%d",&t); while(t--) { scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d%d",&s[i].length,&s[i].weight); s[i].vis = false; } sort(s,s+n,cmp); int ans = 0; for(int i=0;i<n;i++) { if(!s[i].vis) { s[i].vis=true; ++ans; int weight = s[i].weight; for(int j=i+1;j<n;j++) { if(!s[j].vis && s[j].weight>=weight) { s[j].vis = true; weight = s[j].weight; } } } } printf("%d\n",ans); } return 0; }