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. ELF--动态链接

    对前面add.c稍作修改, #include <stdio.h>int add_count = 0; extern int sum_count;extern void print_log( ...

  2. Arrays, Hashtables and Dictionaries

    Original article Built-in arrays Javascript Arrays(Javascript only) ArrayLists Hashtables Generic Li ...

  3. UI:基础

    App的生命周期 参考 多态的使用 // // main.m #import <Foundation/Foundation.h> #import "SingleDog.h&quo ...

  4. phantomjs

    PhantomJS是以WebKit为核心并提供JavaScript编程接口(API)的无界面浏览器. 它提供对web标准的 快速 并且 原生 的支持: DOM操作.CSS选择符.JSON.Canvas ...

  5. 浅析WINFORM工具条的重用实现

    一直以来,我都想看看别人家的工具栏重用(图1)到底是如何实现的,但在网上搜索了很久都没有找到过,即使找到一些程序,要么就是把这个工具栏写在具体的画面(图2),要么就是没有源代码的, 我在想,是否别人也 ...

  6. Commons IO方便读写文件的工具类

    Commons IO是apache的一个开源的工具包,封装了IO操作的相关类,使用Commons IO可以很方便的读写文件,url源代码等. 普通地读取一个网页的源代码的代码可能如下 InputStr ...

  7. jackson 解析json问题

    1.json串中有key为A,但指定转换的mybean中未定义属性A,会抛异常.处理:mapper.configure(Feature.FAIL_ON_UNKNOWN_PROPERTIES, fals ...

  8. boost 线程、互斥体、条件变量

    1.任何技术都是针对特定场景设计的,也就是说,为了解决某个问题而设计的. 2.考虑下面一种场景:一个小旅馆,只有一个卫生间,有清洁人员,店主人,和旅客.卫生间用完之后,就会自动锁闭,必须取钥匙,才能进 ...

  9. easy Html5 - Jquery Mobile之ToolBars(Header and Footer)

    jquery 在web js框架上的风暴还在继续却也随着移动终端走向了mobile:那么jquery mobile到底包括些什么呢 简介工具栏是在移动网站和应用中的头部,尾部或者内容中的工具条:Jqu ...

  10. IOS 7 Study - Displaying an Image on a Navigation Bar

    ProblemYou want to display an image instead of text as the title of the current view controlleron th ...