FatMouse's Speed

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Appoint description: 
prayerhgq  (2015-07-28)
System Crawler  (2015-09-05)

Description

FatMouse believes that the fatter a mouse is, the faster it runs. To disprove this, you want to take the data on a collection of mice and put as large a subset of this data as possible into a sequence so that the weights are increasing, but the speeds are decreasing. 
 

Input

Input contains data for a bunch of mice, one mouse per line, terminated by end of file.

The data for a particular mouse will consist of a pair of integers: the first representing its size in grams and the second representing its speed in centimeters per second. Both integers are between 1 and 10000. The data in each test case will contain information for at most 1000 mice.

Two mice may have the same weight, the same speed, or even the same weight and speed.

 

Output

Your program should output a sequence of lines of data; the first line should contain a number n; the remaining n lines should each contain a single positive integer (each one representing a mouse). If these n integers are m[1], m[2],..., m[n] then it must be the case that

W[m[1]] < W[m[2]] < ... < W[m[n]]

and

S[m[1]] > S[m[2]] > ... > S[m[n]]

In order for the answer to be correct, n should be as large as possible. 
All inequalities are strict: weights must be strictly increasing, and speeds must be strictly decreasing. There may be many correct outputs for a given input, your program only needs to find one. 

 

Sample Input

6008 1300
6000 2100
500 2000
1000 4000
1100 3000
6000 2000
8000 1400
6000 1200
2000 1900
 

Sample Output

4
4
5
9
7
 
 
 #include <iostream>
#include <cstdio>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <climits>
using namespace std; const int SIZE = ;
struct Node
{
int pos;
int weight,speed;
int front,num;
}DP[SIZE]; bool comp(const Node & r_1,const Node & r_2);
int main(void)
{
int count = ; while(scanf("%d%d",&DP[count].weight,&DP[count].speed) != EOF)
{
DP[count].pos = count;
DP[count].num = ;
count ++;
}
sort(DP,DP + count,comp); int max = ,max_loc = ;
for(int i = ;i < count;i ++)
{
DP[i].front = i;
for(int j = ;j < i;j ++)
if(DP[i].weight > DP[j].weight && DP[i].speed < DP[j].speed)
if(DP[i].num < DP[j].num)
{
DP[i].num = DP[j].num;
DP[i].front = j;
if(max < DP[i].num + )
{
max = DP[i].num + ;
max_loc = i;
}
}
DP[i].num ++;
} int temp[SIZE];
count = ;
while()
{
temp[count] = max_loc;
if(max_loc == DP[max_loc].front)
{
count ++;
break;
}
max_loc = DP[max_loc].front;
count ++;
}
printf("%d\n",max);
for(int i = count - ;i >= ;i --)
printf("%d\n",DP[temp[i]].pos + ); return ;
} bool comp(const Node & r_1,const Node & r_2)
{
return r_1.weight < r_2.weight;
}

怒刷DP之 HDU 1160的更多相关文章

  1. 怒刷DP之 HDU 1257

    最少拦截系统 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Statu ...

  2. 怒刷DP之 HDU 1260

    Tickets Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Stat ...

  3. 怒刷DP之 HDU 1176

    免费馅饼 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status  ...

  4. 怒刷DP之 HDU 1087

    Super Jumping! Jumping! Jumping! Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64 ...

  5. 怒刷DP之 HDU 1114

    Piggy-Bank Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit S ...

  6. 怒刷DP之 HDU 1069

    Monkey and Banana Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u S ...

  7. 怒刷DP之 HDU 1024

    Max Sum Plus Plus Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u S ...

  8. 怒刷DP之 HDU 1029

    Ignatius and the Princess IV Time Limit:1000MS     Memory Limit:32767KB     64bit IO Format:%I64d &a ...

  9. HDU 1160 DP最长子序列

    G - FatMouse's Speed Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64 ...

随机推荐

  1. android 简易时间轴(实质是ListView)

    ListView的应用 1.在很多时候是要用到时间轴的,有些处理的时间轴比较复杂,这里就给出一个比较简单的时间轴,其实就是ListView里面的Item的设计. 直接上代码: ListView,ite ...

  2. ADO与ADO.NET的区别

    ADO是使用ole db接口并基于微软的COM技术,ADO.NET使用自己的ADO.NET接口并基于微软的.NET体系架构,所以ADO.NET与ADO是两种数据访问方式. ADO以recordset存 ...

  3. JavaScript 各种遍历方式详解,有你不知道的黑科技

    http://segmentfault.com/a/1190000003968126 为了方便例子讲解,现有数组和json对象如下 var demoArr = ['Javascript', 'Gulp ...

  4. sudo权限集中管理用法

    #定义一组命令集合,名称DBA_CMD,禁止使用的命令前加!即可Cmnd_Alias DBA_CMD =  /bin/touch,/bin/mkdir,/sbin/service,/sbin/chkc ...

  5. 最长公共子序列(LCS问题)

    先简单介绍下什么是最长公共子序列问题,其实问题很直白,假设两个序列X,Y,X的值是ACBDDCB,Y的值是BBDC,那么XY的最长公共子序列就是BDC.这里解决的问题就是需要一种算法可以快速的计算出这 ...

  6. 【机器学习算法-python实现】决策树-Decision tree(1) 信息熵划分数据集

    (转载请注明出处:http://blog.csdn.net/buptgshengod) 1.背景 决策书算法是一种逼近离散数值的分类算法,思路比較简单,并且准确率较高.国际权威的学术组织,数据挖掘国际 ...

  7. Approaching the Fun Factor in Game Design

    I recently did some research on this and talked to Dr. Clayton Lewis (computer Scientist in Residenc ...

  8. 007_尚学堂_高明鑫_android 之项目的打包apk与apk的反编译

    http://www.tudou.com/programs/view/kMQlxff8evM/

  9. [AngularJS] Accessing Scope from The Console

    Using Angular, you can actually access the scope and other things from the console, so when you have ...

  10. iOS开发——UI篇Swift篇&UIImageView

    UIImageView override func viewDidLoad() { super.viewDidLoad() titleLabel.text = titleString //通过坐标和大 ...