FatMouse's Speed (hdu 1160)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std; #define met(a,b) (memset(a,b,sizeof(a)))
#define N 1100
#define INF 0xffffff
struct node
{
int w, s, num;
bool operator < (const node &n)const
{
if(n.w!=w)
return w < n.w;
return s > n.s;
}
}m[N]; int dp[N], pre[N];
int a[N]; int main()
{
int i, j, k=, n, Max=, Index; met(m, );
met(pre, );
met(dp, ); while(scanf("%d%d", &m[k].w, &m[k].s)!=EOF)
{
m[k].num=k;
k++;
} sort(m+, m+k+); n = k-; for(i=; i<=n; i++)
{
dp[i] = ;
for(j=; j<i; j++)
{
if(m[j].w<m[i].w && m[i].s<m[j].s)
{
if(dp[i]<dp[j]+)
{
dp[i] = dp[j] + ;
pre[i] = j;
}
}
if(dp[i]>Max)
{
Max = dp[i];
Index = i;
}
}
} k=;
while(Index)
{
a[k++] = m[Index].num;
Index = pre[Index];
} printf("%d\n", Max);
for(i=k-; i>=; i--)
printf("%d\n", a[i]); return ;
}
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.
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.
FatMouse's Speed (hdu 1160)的更多相关文章
- 动态规划----FatMouse’s Speed(HDU 1160)
参考:https://blog.csdn.net/u012655441/article/details/64920825 https://blog.csdn.net/wy19910326/articl ...
- FatMouse's Speed(HDU LIS)
FatMouse's Speed Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
- FatMouse's Speed ~(基础DP)打印路径的上升子序列
FatMouse believes that the fatter a mouse is, the faster it runs. To disprove this, you want to take ...
- (记忆化搜索) FatMouse and Cheese(hdu 1078)
题目大意: 给n*n地图,老鼠初始位置在(0,0),它每次行走要么横着走要么竖着走,每次最多可以走出k个单位长度,且落脚点的权值必须比上一个落脚点的权值大,求最终可以获得的最大权值 (题目很容 ...
- 2道acm编程题(2014):1.编写一个浏览器输入输出(hdu acm1088);2.encoding(hdu1020)
//1088(参考博客:http://blog.csdn.net/libin56842/article/details/8950688)//1.编写一个浏览器输入输出(hdu acm1088)://思 ...
- HDU 1160 FatMouse's Speed(要记录路径的二维LIS)
FatMouse's Speed Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
- HDU 1160:FatMouse's Speed(LIS+记录路径)
FatMouse's Speed Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
- HDU 1160 FatMouse's Speed (动态规划、最长下降子序列)
FatMouse's Speed Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
- Bestcoder13 1003.Find Sequence(hdu 5064) 解题报告
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5064 题目意思:给出n个数:a1, a2, ..., an,然后需要从中找出一个最长的序列 b1, b ...
随机推荐
- JQuery.validate 错误信息对话框
<script src="${pageContext.request.contextPath}/static/js/jquery-1.12.1.js"type="t ...
- 22.Mysql磁盘I/O
22.磁盘I/O问题磁盘IO是数据库性能瓶颈,一般优化是通过减少或延缓磁盘读写来减轻磁盘IO的压力及其对性能的影响.增强磁盘读写性能和吞吐量也是重要的优化手段. 22.1 使用磁盘阵列 RAID(Re ...
- get(0).tagName获得作用标签
<script type="text/javascript" src="jquery1.4.js"></script><scrip ...
- java中的内存模型
概述 Java平台自动集成了线程以及多处理器技术,这种集成程度比Java以前诞生的计算机语言要厉害很多,该语言针对多种异构平台的平台独立性而使用的多线程技术支持也是具有开拓性的一面,有时候在开发Jav ...
- UOJ 274 温暖会指引我们前进 - LCT
Solution 更新掉路径上温暖度最小的边就可以了~ Code #include<cstdio> #include<cstring> #include<algorith ...
- Python之路(第十一篇)装饰器
一.什么是装饰器? 装饰器他人的器具,本身可以是任意可调用对象,被装饰者也可以是任意可调用对象. 强调装饰器的原则:1 不修改被装饰对象的源代码 2 不修改被装饰对象的调用方式 装饰器的目标:在遵循1 ...
- Python3实战系列之三(获取印度售后数据项目)
问题:续接上一篇.说干咱就干呀,勤勤恳恳写程序呀! 目标:实现第一个python程序的“Hello world!” 解决方案:新建一个项目Test,创建一个Test.py文件.在文件中实现打印出“He ...
- fetch获取json的正确姿势
fetch要求参数传递,遇到请求无法正常获取数据,网上其他很多版本类似这样: fetch(url ,{ method: 'POST', headers:{ 'Accept': 'application ...
- OneZero第三周第二次站立会议(2016.4.5)
1. 时间: 13:00--13:15 共计15分钟. 2. 成员: X 夏一鸣 * 组长 (博客:http://www.cnblogs.com/xiaym896/), G 郭又铭 (博客:http ...
- 【WebService】WebService之CXF的拦截器(五)
CXF拦截器介绍 CXF拦截器是功能的主要实现单元,也是主要的扩展点,可以在不对核心模块进行修改的情况下,动态添加功能.当服务被调用时,会经过多个拦截器链(Interceptor Chain)处理,拦 ...