FatMouse's Speed

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15801    Accepted Submission(s): 6969
Special Judge

Problem 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
 

Source

 
 //2017-04-04
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm> using namespace std; struct node
{
int w, s, pos;
bool operator<(node x)
{
return w < x.w;
}
}mice[];
int dp[], pre[];//dp[i]表示前i只老鼠的最长下降子序列 void print(int pos)
{
if(pos == -)return ;
print(pre[pos]);
printf("%d\n", mice[pos].pos);
} int main()
{
int n = , w, s;
memset(pre, -, sizeof(pre));
while(scanf("%d%d", &w, &s)!=EOF)
{
mice[n].w = w;
mice[n].s = s;
mice[n].pos = n+;
n++;
}
sort(mice, mice+n);
int pos = -, mx = ;
for(int i = ; i < n; i++){
dp[i] = ;
for(int j = ; j < i; j++){
if(mice[j].w < mice[i].w && mice[j].s > mice[i].s){
if(dp[j]+ > dp[i]){
dp[i] = dp[j]+;
pre[i] = j;
}
}
}
if(dp[i] > mx){
mx = dp[i];
pos = i;
}
}
printf("%d\n", mx);
print(pos);
return ;
}

HDU1160(KB12-J DP)的更多相关文章

  1. 2016-2017 ACM-ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred) J dp 背包

    J. Bottles time limit per test 2 seconds memory limit per test 512 megabytes input standard input ou ...

  2. J Dp

    <span style="color:#000099;">/* ____________________________________________________ ...

  3. C. Multiplicity 简单数论+dp(dp[i][j]=dp[i-1][j-1]+dp[i-1][j] 前面序列要满足才能构成后面序列)+sort

    题意:给出n 个数 的序列 问 从n个数删去任意个数  删去的数后的序列b1 b2 b3 ......bk  k|bk 思路: 这种题目都有一个特性 就是取到bk 的时候 需要前面有个bk-1的序列前 ...

  4. Andrew Stankevich's Contest (21) J dp+组合数

    坑爹的,,组合数模板,,, 6132 njczy2010 1412 Accepted 5572 MS 50620 KB C++ 1844 B 2014-10-02 21:41:15 J - 2-3 T ...

  5. 牛客集训第七场J /// DP

    题目大意: 在矩阵(只有52种字符)中找出所有不包含重复字符的子矩阵个数 #include <bits/stdc++.h> #define ll long long using names ...

  6. hdu1160 dp

    hdu1160 题意:给出很多老鼠的数据,分别是它们的体重和跑速,为了证明老鼠越重跑得越慢,要找一组数据,由若干个老鼠组成,保证老鼠的体重依次增加而跑速依次减小,问这组数据最多能有多少老鼠,并按体重从 ...

  7. hdu 4049 2011北京赛区网络赛J 状压dp ***

    cl少用在for循环里 #include<cstdio> #include<iostream> #include<algorithm> #include<cs ...

  8. 2017 ICPC区域赛(西安站)--- J题 LOL(DP)

    题目链接 problem description 5 friends play LOL together . Every one should BAN one character and PICK o ...

  9. codeforces gym 100947 J. Killing everything dp+二分

    J. Killing everything time limit per test 4 seconds memory limit per test 64 megabytes input standar ...

随机推荐

  1. git使用分支与tag

    查看分支:git branch 创建分支:git branch <name> 切换分支:git checkout <name> 创建+切换分支:git checkout -b ...

  2. PowerDesigner生成OOM时类名属性名转换

    Examples Script 1: Convert a name into a class code (JAVA naming convention)转换类名 .foreach_part(%Name ...

  3. TCP连接详解

    一. 连接过程示意图 二. 建立TCP连接 2.1 三次握手 第一次握手:建立连接.客户端发送连接请求报文段,将SYN置为1,Sequence Number为x:然后,客户端进入SYN_SEND状态, ...

  4. javap 反汇编class文件

    用法: javap 参数 class文件路径 其中, 可能的选项包括: -help --help -? 输出此用法消息 -version 版本信息 -v -verbose 输出附加信息 -l 输出行号 ...

  5. @transactional作用和事务

    今天在博客园看到有发布spring的注解,留意到@transactional这个注解.立马就百度.学习了 使用这个注解的类或者方法表示该类里面的所有方法或者这个方法的事务由spring处理,来保证事务 ...

  6. flask_ Mongodb 的语法-排序

    MOngoDB的排序是挺有用的   ,跟MySQL有明显的区别 .. 它的原生语法的第一个参数为条件限定,第二个参数为排序字段 db.news.find({},{'_id':1})   #1是升序  ...

  7. Python Web框架 tornado 异步原理

    Python Web框架 tornado 异步原理 参考:http://www.jb51.net/article/64747.htm 待整理

  8. spring security xml配置详解

    security 3.x <?xml version="1.0" encoding="UTF-8"?> <beans:beans xmlns= ...

  9. pipelinedb--流、滑动窗口测试

    https://blog.csdn.net/liuxiangke0210/article/details/74010951 https://yq.aliyun.com/articles/166 一.p ...

  10. mongorestore 一次踩雷

    1.在做mongodb备份后,研发突然有个需求说先看一下昨天备份里面的数据,进行一下核实.因为那部分数据今天已经删除,由于使用---gzip.--archive做的备份,所以必须导入到同名的数据库里面 ...