http://acm.hdu.edu.cn/showproblem.php?pid=1051

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
 
题解:贪心 先排序
时间复杂度:$O(N ^ 2 + N * logN)$
代码:

#include <bits/stdc++.h>
using namespace std; const int maxn = 1e5 + 10;
int N; struct Node {
int l;
int w;
int flag;
}node[maxn]; bool cmp(const Node& a, const Node& b) {
return a.l == b.l ? a.w < b.w : a.l < b.l;
} int main() {
int T;
scanf("%d", &T);
while(T --) {
memset(node, 0, sizeof(node));
scanf("%d", &N);
for(int i = 1; i <= N; i ++) {
scanf("%d%d", &node[i].l, &node[i].w);
node[i].flag = 0;
} int cnt = 0;
sort(node + 1, node + 1 + N, cmp);
for (int k = 1; k <= N;)
{
cnt ++;
int L = 0, W = 0;
for (int i = 1; i <= N; i ++) {
if (!node[i].flag)
if (node[i].l >= L && node[i].w >= W) {
node[i].flag = 1;
L = node[i].l;
W = node[i].w;
k ++;
}
}
}
printf("%d\n", cnt);
}
return 0;
}

  

HDU 1005 Wooden Sticks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||DP

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

  3. 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)    ...

  4. 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 ...

  5. 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 ...

  6. HDU 1051 Wooden Sticks【LIS】

    题意:给出n个木头的重量wi,长度li,如果满足w[i+1]>=w[i]且l[i+1]>=l[i],则不用耗费另外的加工时间,问至少需要多长时间加工完这些木头. 第一次做这一题目也没有做出 ...

  7. 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 ...

  8. 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 ...

  9. hdu 1051 wooden sticks (贪心+巧妙转化)

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

随机推荐

  1. 自定义组件v-model的实质性理解

    用了几个月Vue一直很纠结自定义组件的v-model实现,最近开始学习React时,React中受控组件与状态提升的理念与v-model不谋而合. 转载请注明地址: https://www.cnblo ...

  2. 第13届景驰-埃森哲杯广东工业大学ACM程序设计大赛--F-等式

    链接:https://www.nowcoder.com/acm/contest/90/F 来源:牛客网 1.题目描述 给定n,求1/x + 1/y = 1/n (x<=y)的解数.(x.y.n均 ...

  3. 【TOJ 1545】Hurdles of 110m(动态规划)

    描述 In the year 2008, the 29th Olympic Games will be held in Beijing. This will signify the prosperit ...

  4. ABAP术语-Call Transaction

    Call Transaction 原文:http://www.cnblogs.com/qiangsheng/archive/2008/01/15/1039270.html A data transfe ...

  5. JVM 垃圾回收机制和常见算法和 JVM 的内存结构和内存分配(面试题)

    一.JVM 垃圾回收机制和常见算法 Sun 公司只定义了垃圾回收机制规则而不局限于其实现算法,因此不同厂商生产的虚拟机采用的算法也不尽相同.GC(Garbage Collector)在回收对象前首先必 ...

  6. VULTR的VPS在centos的操作系统中出现网站无法访问 80端口被firewall禁止

    导语:叶子在为一位客户配置web服务器环境的时候,出现网站不能访问的情况,但ping正常.客户的服务器是在VULTR上购买的VPS,安装的操作系统为centos 7.3.经过叶子的分析,认为是防火墙阻 ...

  7. DevOps - 配置管理 - Ansible

    http://www.zsythink.net/archives/category/运维相关/ansible/

  8. 序列化serialize()与反序列化unserialize()的实例

    在写序列化serialize与反序列化unserialize()时,我们先来看看: serialize - 产生一个可存储的值的表示 描述 string serialize ( mixed $valu ...

  9. mongodb导入全栈商城的goods和users数据

    > show dbsshow dbsadmin 0.000GBconfig 0.000GBlocal 0.000GB> use dumalluse dumallswitched to db ...

  10. over开窗函数的用法

    over(partition by c1.pmid,d1.type,e1.objid  order by e1.objid ) pinum 先根据字段排序,pinum.在取第一条数据and p1.pi ...