先将火柴按照长度(或重量)优先排序,在不断遍历数组,找出其中重量(长度)递增子序列,并标记


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
#include<iostream>
#include<algorithm>
using namespace std;
struct wood{
int l;
int w;
int flag;
}W[];
bool cmp(wood w1, wood w2){
if(w1.l<w2.l )
return true;
else if(w1.l==w2.l && w1.w<=w2.w)
return true;
return false;
}
int main()
{
int t,n;
cin>>t;
while(t--){
cin>>n;
for(int i=;i<n;i++){
cin>>W[i].l>>W[i].w;
W[i].flag=;
}
sort(W,W+n,cmp);
//for(int i=0;i<n;i++)
//cout<<W[i].l<<" "<<W[i].w<<endl;
int tw;
int result=;
for(int i=;i<n;i++){
if(W[i].flag)
continue;
tw=W[i].w;
W[i].flag=;
for(int j=;j<n;j++){
if(W[j].flag)
continue;
if(W[j].w>=tw){
tw=W[j].w;
W[j].flag=;
}
}
result++;
}
cout<<result<<endl;
}
return ;
}
 

贪心-Wooden Sticks的更多相关文章

  1. 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 ...

  2. HDOJ-1051 Wooden sticks(贪心)

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

  3. POJ 1065 Wooden Sticks (贪心)

    There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The st ...

  4. 1270: Wooden Sticks [贪心]

    点击打开链接 1270: Wooden Sticks [贪心] 时间限制: 1 Sec 内存限制: 128 MB 提交: 31 解决: 11 统计 题目描述 Lialosiu要制作木棍,给n根作为原料 ...

  5. 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 ...

  6. 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 ...

  7. HDU 1051 Wooden Sticks (贪心)

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

  8. uvalive 2322 Wooden Sticks(贪心)

    题目连接:2322 Wooden Sticks 题目大意:给出要求切的n个小木棍 , 每个小木棍有长度和重量,因为当要切的长度和重量分别大于前面一个的长度和重量的时候可以不用调整大木棍直接切割, 否则 ...

  9. HDU 1051 Wooden Sticks 贪心||DP

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

随机推荐

  1. python运行时参数m的作用

    不加m时,当前目录是py文件的所在目录 加m时,当前目录就是当前目录

  2. java 深度复制与浅复制 copyOf、arraycopy、copyOfRange

    1.copyOf 原型:public static <T> T[] copyOf(T[] original, int newLength) original:原数组 newLength:要 ...

  3. MySQL 权限相关

    # ============================= mysql 权限相关 =====================================================gran ...

  4. C# 打印 长字符串自动换行

    主要代码如下: StringFormat fmt = new StringFormat(); fmt.LineAlignment = StringAlignment.Near;//左对齐 fmt.Fo ...

  5. EF 数据迁移 常见错误

    1.错误 “LC.exe”已退出,代码为 -1 原因:解决方案出错,而非迁移的项目

  6. python day07笔记总结

    2019.4.4 S21  day07笔记总结 一.深浅拷贝 1.copy.copy()     浅拷贝 deep.copy()    深拷贝 2.一般情况 1.str/int/bool 是不可变类型 ...

  7. scrapy 爬取豆瓣互联网图书

    安装scrapy conda install scrapy 生成一个scrapy项目 scrapy startproject douban settings文件 # -*- coding: utf-8 ...

  8. Linux - Ubuntu 图形界面入门

    Ubuntu 图形界面入门 目标 熟悉 Ubuntu 图形界面的基本使用 01. Ubuntu 的任务栏 02. 窗口操作按钮 03. 窗口菜单条 ——本文源自<黑马程序员>

  9. 【相关网站 - 01】- Java 相关网站

    一.官方网站 1. Java 官方网站 https://www.java.com/zh_CN/ 2. Spring 官方网站 http://spring.io/ 1. Spring Framework ...

  10. fail-fast和fail-safe的区别

    fail-fast(快速失败):多线程情况下,一个线程通过迭代器读取集合中的值时,另一个线程修改了集合,则会抛出ConcurrentModificationException异常: 集合中通过modC ...