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
- PAT 1051 复数乘法(15 )(代码+思路)
1051 复数乘法(15 分) 复数可以写成 (A+Bi) 的常规形式,其中 A 是实部,B 是虚部,i 是虚数单位,满足 i2=−1:也可以写成极坐标下的指数形式 (R×e(Pi)),其 ...
- Codeforces 799D. String Game 二分
D. String Game time limit per test:2 seconds memory limit per test:512 megabytes input:standard inpu ...
- com_pc-mcu
#include <REG52.H> unsigned char UART_buff; bit New_rec = , Send_ed = , Money = ; void main (v ...
- 高负载PHP调优
高负载PHP调优 针对PHP的Linux调优 调整文件描述符限制 # ulimit -n 1000000 # vi /etc/security/limits.conf # Setting Shell ...
- 【SpringAop】【统一日志处理】注解方式理解以及使用
[注意:本次代码的demo会存在百度网盘,由于公司的保密,禁止上传,所以仅本人可见] 目前公司在做数据资产项目,数据质量部分使用到了springaop做统一日志处理,以前对这块有了解,有点模糊不清,今 ...
- 句子相似度_tf/idf
import mathfrom math import isnanimport pandas as pd#结巴分词,切开之后,有分隔符def jieba_function(sent): import ...
- mysql索引相关
索引有主键索引.唯一索引.普通索引 单列索引,复合索引. 复合索引(a,b,c),可以理解是有三个索引,分别是a.b.c三个索引 前缀不是a的话,复合索引都不起作用,前缀用函数或者是范围,比如< ...
- mysql 压缩版安装
环境介绍:win2008_x64+mysql5.7.10 64位 1.将压缩包解压到d:\\mysql目录,并将mysql目录中的my-default.ini 重命名为my.ini 2.将my.in ...
- 通过SD卡来安装Linux系统
一.制作SD启动卡(安装Linux)步骤: 烧写原理:superboot-6410.bin(bootloader)+内核镜像文件+根文件系统 1.将SD卡插入USB接口的读卡器,并插在PC的USB口 ...
- oracle 62进制序列号
create or replace function GetSerial62(v_lpad number default 0) return varchar2 IS v_tmp number(38,0 ...