Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22000    Accepted Submission(s): 8851

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
思路:将木棒按l,w 递减的顺序排列,统计关于递减的连续子列的个数即为答案。
代码:

 #include "cstdio"
 #include "iostream"
 #include "algorithm"
 #include "string"
 #include "cstring"
 #include "queue"
 #include "cmath"
 #include "vector"
 #include "map"
 #include "stdlib.h"
 #include "set"
 #define mj
 #define db double
 #define ll long long
 using  namespace std;
 ;
 ;
 ;
 bool u[N];
 typedef  struct stu{
     int l,w;
 }M;
 M s[];
 int cmp(M a,M b)
 {
     if(a.l==b.l) return a.w>=b.w;
     return a.l>=b.l;
 }
 int main()
 {
     int t,n,i,j,ans;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&n);
         ;i<n;i++)
             scanf(;
         sort(s,s+n,cmp);
         int mi;
         ans=;
         ;i<n;i++)
         {
             if(u[i]) continue;
             mi=s[i].w;
             ;j<n;j++)
             {
                 if(!u[j]&&s[j].w<=mi){
                     mi=s[j].w;
                     u[j]=;
                 }
             }
             ans++;
         }
         printf("%d\n",ans);
     }
     ;
 }
 

HDU 1051 Wooden Sticks 贪心||DP的更多相关文章

  1. HDU 1051 Wooden Sticks (贪心)

    Wooden Sticks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) To ...

  2. HDU - 1051 Wooden Sticks 贪心 动态规划

    Wooden Sticks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)    ...

  3. HDU 1051 Wooden Sticks 贪心题解

    本题一看就知道是最长不减序列了,一想就以为是使用dp攻克了. 只是那是个错误的思路. 我就动了半天没动出来.然后看了看别人是能够使用dp的,只是那个比較难证明其正确性,而其速度也不快.故此并非非常好的 ...

  4. hdu 1051 wooden sticks (贪心+巧妙转化)

    #include <iostream>#include<stdio.h>#include<cmath>#include<algorithm>using ...

  5. hdu 1051:Wooden Sticks(水题,贪心)

    Wooden Sticks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Tot ...

  6. HDOJ 1051. Wooden Sticks 贪心 结构体排序

    Wooden Sticks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) To ...

  7. HDOJ.1051 Wooden Sticks (贪心)

    Wooden Sticks 点我挑战题目 题意分析 给出T组数据,每组数据有n对数,分别代表每个木棍的长度l和重量w.第一个木棍加工需要1min的准备准备时间,对于刚刚经加工过的木棍,如果接下来的木棍 ...

  8. HDU 1051 Wooden Sticks 造木棍【贪心】

    题目链接>>> 转载于:https://www.cnblogs.com/Action-/archive/2012/07/03/2574800.html  题目大意: 给n根木棍的长度 ...

  9. HDU 1051 Wooden Sticks

    题意: 有 n 根木棒,长度和质量都已经知道,需要一个机器一根一根地处理这些木棒. 该机器在加工过程中需要一定的准备时间,是用于清洗机器,调整工具和模板的. 机器需要的准备时间如下: 1.第一根需要1 ...

随机推荐

  1. 分布式版本控制git常见问题之gitignore冲突(精简版)

    上次写的的太模糊了,现在简单直接写出个人心得,如下: 原因是有人提交了.gitignore里面的内容,所以和本地的不一样,这样就有问题,那么pull都不可以,所以要这样: git update-ind ...

  2. 【面向对象设计原则】之接口隔离原则(ISP)

    接口隔离原则(Interface  Segregation Principle, ISP):使用多个专门的接口,而不使用单一的总接口,即客户端不应该依赖那些它不需要的接口. 从接口隔离原则的定义可以看 ...

  3. 【JAVAWEB学习笔记】网上商城实战5:后台的功能模块

    今日任务 完成后台的功能模块 1.1      网上商城的后台功能的实现: 1.1.1    后台的功能的需求: 1.1.1.1  分类管理: [查询所有分类] * 在左侧菜单页面中点击分类管理: * ...

  4. MySQL存储汉字

    之前在网上查找了很多方法,排在前排的都是修改配置文件my.ini的,没有成功,后来找到了一个解决方法: 在建表的时候,在语句后面加上段"engine = innodb default cha ...

  5. Unable to find 'struts.multipart.saveDir' property setting.

    今天在项目开发中遇到如下问题 项目使用的是struts2 Unable to find 'struts.multipart.saveDir' property setting. 后来在网上查询特此记录 ...

  6. Java NIO学习笔记六 SocketChannel 和 ServerSocketChannel

    Java NIO SocketChannel Java NIO SocketChannel是连接到TCP网络socket(套接字)的通道.Java NIO相当于Java Networking的sock ...

  7. 原生js二级联动

    今天说的这个是原生js的二级联动,在空白页面里动态添加并作出相对应的效果. 1 //创建两个下拉列表 select标签 是下拉列表 var sel = document.createElement(& ...

  8. 题解 最优的挤奶方案(Optimal Milking)

    最优的挤奶方案(Optimal Milking) 时间限制: 1 Sec  内存限制: 128 MB 题目描述 农场主 John 将他的 K(1≤K≤30)个挤奶器运到牧场,在那里有 C(1≤C≤20 ...

  9. web前端面试总结(二)

    这段时间大大小小面试确实不少,相对之前那篇被虐到体无完肤这几次确实相对来说有很大进步这里总结一下: 1.发现自己,站在个人角度我还是挺赞成出去面试的,不管你对现在的公司是否满意,当你觉得在这里已经有一 ...

  10. svn命令行便捷代码

    在把分支merge回主干的时候,有时候需要只提交自己修改过的文件,但是很多文件其实分支上没动过,但却显示有变化,这个其实是属性发生了变化.svn通过svn:mergeinfo来记录merge的记录.所 ...