Some people think that the bigger an elephant is, the smarter it is. To disprove this, you want to take the data on a collection of elephants and put as large a subset of this data as possible into a sequence so
that the weights are increasing, but the IQ's are decreasing.

The input will consist of data for a bunch of elephants, one elephant per line, terminated by the end-of-file. The data for a particular elephant will consist of a pair of integers: the first representing its size
in kilograms and the second representing its IQ in hundredths of IQ points. Both integers are between 1 and 10000. The data will contain information for at most 1000 elephants. Two elephants may have the same weight, the same IQ, or even the same weight and
IQ.

Say that the numbers on the i-th data line are W[i] and S[i]. 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 an elephant). If these n integers are a[1]a[2],..., a[n] then it must be the case that

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

and

   S[a[1]] > S[a[2]] > ... > S[a[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
IQs 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<cstring>
#include<algorithm>
#include<limits.h>
typedef long long LL;
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ;2 i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
struct node{
int id;
int w,s;
}e[1100];
int dp[1100];
int pre[1100];
int ans=-1,pos;
int cmp(node l1,node l2)
{
return l1.w<l2.w;
}
void print(int x)
{
if(pre[x]==-1)
{
printf("%d\n",e[x].id);
return ;
}
print(pre[x]);
printf("%d\n",e[x].id);
}
int main()
{
int num=1;
while(~scanf("%d%d",&e[num].w,&e[num].s))
{
e[num].id=num;
num++;
}
sort(e+1,e+num,cmp);
ans=0;CLEAR(pre,-1);
for(int i=1;i<=num;i++)
{
dp[i]=1;
for(int j=1;j<i;j++)
{
if(e[i].w>e[j].w&&e[i].s<e[j].s&&dp[i]<dp[j]+1)
{
dp[i]=dp[j]+1;
pre[i]=j;
}
if(dp[i]>ans)
{
ans=dp[i];
pos=i;
}
}
}
printf("%d\n",ans);
print(pos);
return 0;
}

UVA 10131 Is Bigger Smarter?(DP)的更多相关文章

  1. uva 10131 Is Bigger Smarter?(DAG最长路)

    题目连接:10131 - Is Bigger Smarter? 题目大意:给出n只大象的属性, 包括重量w, 智商s, 现在要求找到一个连续的序列, 要求每只大象的重量比前一只的大, 智商却要小, 输 ...

  2. UVA 10131 Is Bigger Smarter?(DP最长上升子序列)

    Description   Question 1: Is Bigger Smarter? The Problem Some people think that the bigger an elepha ...

  3. UVA 10131 - Is Bigger Smarter? (动态规划)

    Is Bigger Smarter? The Problem Some people think that the bigger an elephant is, the smarter it is. ...

  4. Uva 10131 Is Bigger Smarter? (LIS,打印路径)

    option=com_onlinejudge&Itemid=8&page=show_problem&problem=1072">链接:UVa 10131 题意: ...

  5. uva 10131 Is Bigger Smarter ? (简单dp 最长上升子序列变形 路径输出)

    题目链接 题意:有好多行,每行两个数字,代表大象的体重和智商,求大象体重越来越大,智商越来越低的最长序列,并输出. 思路:先排一下序,再按照最长上升子序列计算就行. 还有注意输入, 刚开始我是这样输入 ...

  6. UVa 10131: Is Bigger Smarter?

    动态规划题.类似UVa103 Stacking Box,都是题目给一种判断嵌套的方法然后求最长序列.提前对数据排序可以节省一些时间开销. 我的解题代码如下: #include <iostream ...

  7. UVA - 10131Is Bigger Smarter?(DAG上的DP)

    题目:UVA - 10131Is Bigger Smarter? (DAG) 题目大意:给出一群大象的体重和IQ.要求挑选最多的大象,组成一个序列.严格的体重递增,IQ递减的序列.输出最多的大象数目和 ...

  8. UVA 10131题解

    第一次写动态规划的代码,整了一天,终于AC. 题目: Question 1: Is Bigger Smarter? The Problem Some people think that the big ...

  9. UVA.10066 The Twin Towers (DP LCS)

    UVA.10066 The Twin Towers (DP LCS) 题意分析 有2座塔,分别由不同长度的石块组成.现在要求移走一些石块,使得这2座塔的高度相同,求高度最大是多少. 问题的实质可以转化 ...

随机推荐

  1. Jconsole: JAVA 监视和管理控制台简介

    Jconsole: JAVA 监视和管理控制台简介 JDK中除了提供大量的命令行之外,还提供两个功能强大的可视化工具:JConsole和VisualVM. 之前对java的调试一直停留在 右键-> ...

  2. POJ 3997 Stock Chase

    Stock Chase Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 455   Accepted: 131 Descrip ...

  3. 异常:ERROR [org.hibernate.proxy.BasicLazyInitializer] - CGLIB Enhancement failed...

    ERROR [org.hibernate.proxy.BasicLazyInitializer] - CGLIB Enhancement failed: com.movie.类 放到lib 包下 \W ...

  4. 了解SQL注入攻击

    SQL注入:利用现有应用程序,将(恶意)的SQL命令注入到后台数据库引擎执行的能力,这是SQL注入的标准释义. 随着B/S模式被广泛的应用,用这种模式编写应用程序的程序员也越来越多,但由于开发人员的水 ...

  5. Vanya and Scales(思维)

    Vanya and Scales time limit per test 1 second memory limit per test 256 megabytes input standard inp ...

  6. MTK 2G芯片使用联通卡在深圳无法拨打112原因

    2.75G GSM模块在深圳客户这边联调到最后,客户这边遇到各种概率性问题,基本都是对方使用的配件不够好造成的,如天线.SIM卡座等配件. 一旦这些配件不好,就会出现概率性的错误,非常难以复现,所以在 ...

  7. Cocos2d-x3.1UserDefaule类具体解释

    在Cocos2d-x存储数据使用的类是UserDefault类,以下分析下该类的使用 //.h #include "base/CCPlatformMacros.h" #includ ...

  8. 基于 koajs 的前后端分离实践

    一.什么是前后端分离? 前后端分离的概念和优势在这里不再赘述,有兴趣的同学可以看各个前辈们一系列总结和讨论: 系列文章:前后端分离的思考与实践(1-6) slider: 淘宝前后端分离实践 知乎提问: ...

  9. windows 7 旗舰版 切换 中英文 界面

    http://jingyan.baidu.com/article/f7ff0bfc4963612e26bb131e.html 如果遇到:想下载英语语言包,但是出现代码80070643,windowsu ...

  10. 【IOS学习基础】weak和strong、懒加载、循环引用

    一.weak和strong 1.理解 刚开始学UI的时候,对于weak和strong的描述看得最多的就是“由ARC引入,weak相当于OC中的assign,但是weak用于修饰对象,但是他们都不会造成 ...