D. Dog Show

time limit per test2 seconds

memory limit per test256 megabytes

Problem Description

A new dog show on TV is starting next week. On the show dogs are required to demonstrate bottomless stomach, strategic thinking and self-preservation instinct. You and your dog are invited to compete with other participants and naturally you want to win!

On the show a dog needs to eat as many bowls of dog food as possible (bottomless stomach helps here). Dogs compete separately of each other and the rules are as follows:

At the start of the show the dog and the bowls are located on a line. The dog starts at position x = 0 and n bowls are located at positions x = 1, x = 2, …, x = n. The bowls are numbered from 1 to n from left to right. After the show starts the dog immediately begins to run to the right to the first bowl.

The food inside bowls is not ready for eating at the start because it is too hot (dog’s self-preservation instinct prevents eating). More formally, the dog can eat from the i-th bowl after ti seconds from the start of the show or later.

It takes dog 1 second to move from the position x to the position x + 1. The dog is not allowed to move to the left, the dog runs only to the right with the constant speed 1 distance unit per second. When the dog reaches a bowl (say, the bowl i), the following cases are possible:

  • the food had cooled down (i.e. it passed at least ti seconds from the show start): the dog immediately eats the food and runs to the right without any stop,
  • the food is hot (i.e. it passed less than ti seconds from the show start): the dog has two options: to wait for the i-th bowl, eat the food and continue to run at the moment ti or to skip the i-th bowl and continue to run to the right without any stop.

After T seconds from the start the show ends. If the dog reaches a bowl of food at moment T the dog can not eat it. The show stops before T seconds if the dog had run to the right of the last bowl.

You need to help your dog create a strategy with which the maximum possible number of bowls of food will be eaten in T seconds.

Input

Two integer numbers are given in the first line - n and T (1 ≤ n ≤ 200 000, 1 ≤ T ≤ 2·109) — the number of bowls of food and the time when the dog is stopped.

On the next line numbers t1, t2, …, tn (1 ≤ ti ≤ 109) are given, where ti is the moment of time when the i-th bowl of food is ready for eating.

Output

Output a single integer — the maximum number of bowls of food the dog will be able to eat in T seconds.

Note

In the first example the dog should skip the second bowl to eat from the two bowls (the first and the third).


  • 题意就是一只狗,去吃一列的饭,每次移动到下一个碗花费1秒的时间,瞬间吃完,每个饭的冷却时间为第c[i]秒(只有当饭冷却之后狗才能吃)。问在m秒内这只狗最多能吃多少碗饭。
  • 这是一个经典的贪心问题,狗每次尽可能的多吃饭。就是处理在等待的时间能不能吃更多的饭。枚举第m秒到达的每一个碗。主要是看实现方法。

/*可以把狗想象成可以后悔,狗要将遇到的每一个碗里的饭都吃下去,
当狗发现后面还可以吃但是时间不够了的时候,狗可以把饭吐出来
而老天把狗等待这个饭的时间还给狗,优先队列模拟狗肚子里的饭
*/
#include<bits/stdc++.h>
using namespace std;
int main()
{
priority_queue <int> qu;
int n,m;
scanf("%d%d",&n,&m);
int Max = 0;
for(int i=1;i<=min(m-1,n);i++)
{
int temp;
scanf("%d",&temp);
while(!qu.empty() && qu.top() >= m-i)//每次时间不够的时候去除等待时间最长的那个碗
qu.pop();
if(max(temp,i) < m)
qu.push(temp-i);
Max = max((int)qu.size(),Max);
}
printf("%d",Max);
return 0;
}

CodeForces:847D-Dog Show的更多相关文章

  1. Arduino - 看门狗定时器(WDT:Watch Dog Timer)

    看门狗定时器(WDT:Watch Dog Timer)实际上是一个计数器. 一般给看门狗一个大数,程序开始运行后看门狗开始倒计数. 如果程序运行正常,过一段时间CPU应该发出指令让看门狗复位,令其重新 ...

  2. CodeForces:#448 div2 B. XK Segments

    传送门:http://codeforces.com/contest/895/problem/B B. XK Segments time limit per test1 second memory li ...

  3. CodeForces:#448 div2 a Pizza Separation

    传送门:http://codeforces.com/contest/895/problem/A A. Pizza Separation time limit per test1 second memo ...

  4. YUM:Yellow dog Updater Modified

    1. 什么是YUM YUM(全称为 Yellow dog Updater Modified) 是一个在Fedora和RedHat以及CentOS中的Shell前端软件包管理器.基于RPM包管理,能够从 ...

  5. LCD实验学习笔记(三):WATCH DOG

    看门狗是为了能够防止程序跑飞用的.程序应该定时的去喂狗.如果程序跑飞了,那么就不会去喂狗了.如果超过了喂狗的时间,那么狗就会生成一个信号来reset CPU.一般程序不需要,特殊情况下需要这种机制. ...

  6. Codeforces:68A-Irrational problem(暴力大法好)

    A- Irrational problem p Time Limit: 2000MS Memory Limit: 262144K 64bit IO Format: %I64d& %I64 De ...

  7. CodeForces:148D-D.Bag of mice

    Bag of mice time limit per test 2 seconds memory limit per test 256 megabytes Program Description Th ...

  8. CodeForces:699B-One Bomb

    B. One Bomb time limit per test1 second memory limit per test256 megabytes Problem Description You a ...

  9. SIFT算法:DoG尺度空间生产

    SIFT算法:DoG尺度空间生产  SIFT算法:KeyPoint找寻.定位与优化 SIFT算法:确定特征点方向  SIFT算法:特征描述子 目录: 1.高斯尺度空间(GSS - Gauss Scal ...

随机推荐

  1. A.走方格

    链接:https://ac.nowcoder.com/acm/contest/368/A 题意: 在一个n*n的方格中,你只能斜着走. 你还有一次上下左右走的机会 给你一个起点(sx,sy),和终点( ...

  2. UVA10305:Ordering Tasks(拓扑排序)

    John has n tasks to do. Unfortunately, the tasks are not independent and the execution of one task i ...

  3. override和overload的小笔记

    override是覆盖的意思,也就是我们的重写.可以重写覆盖父类的方法,然后实现接口的方法也可以叫做override. 几个要注意的点: 重写一定要用和被重写方法同样的方法名还有参数列表. 抛出的异常 ...

  4. R 语言中 data table 的相关,内存高效的 增量式 data frame

    面对的是这样一个问题,不断读入一行一行数据,append到data frame上,如果用dataframe,  rbind() ,可以发现数据大的时候效率明显变低. 原因是 每次bind 都是一次重新 ...

  5. Java方式配置Spring MVC

    概述 使用Java方式配置Spring MVC,以及回顾一下Spring MVC的各种用法. Spring MVC简述 关于Spring MVC的介绍网上有很多,这里就不再赘述了,只是要说一下,Spr ...

  6. UEditor的KityFormula在IIS中部署,显示不了的解决方案

    在此,首先感谢我的同事,找到了问题所在. 因Web项目中需要有输入公式的功能(高等数学中需要),普通公式插件无法满足,所以找了KityFormula这款插件. 看了下里面的公式,在数学方面确实比较全面 ...

  7. echarts 添加Loading 等待。

    capturedsDetailsEcharts: function(id) { if (!id) { id = mini.get("chnNameCaptureds").getVa ...

  8. Wrapper class package.jaxws.methodName is not found. Have you run APT to generate them?解决方案

    使用JAX-WS 2.X基于Web容器发布WebService报错,错误信息类似于: Wrapper class package.jaxws.methodName is not found. Have ...

  9. iOS之核心动画

    .将动画的所有方法封装到一个类里面 MyCAHelper.h #import <Foundation/Foundation.h> #import <QuartzCore/Quartz ...

  10. java.lang.UnsatisfiedLinkError: dlopen failed: /data/app/xxx/lib/arm/liblame.so: has text relocations

    最近在写本地录音转码过程中引入了liblame.so,我这边用了不同系统版本的手机测试本地录音都没有出现问题,但是有一天,同事在测试的时候,出现了以下错误: 09-13 17:32:29.140 26 ...