zoukankan      html  css  js  c++  java
  • 校内选拔I题题解 构造题 Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 2) ——D

    http://codeforces.com/contest/574/problem/D

    Bear and Blocks

    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of hi identical blocks. For clarification see picture for the first sample.

    Limak will repeat the following operation till everything is destroyed.

    Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.

    Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.

    Input

    The first line contains single integer n (1 ≤ n ≤ 105).

    The second line contains n space-separated integers h1, h2, ..., hn (1 ≤ hi ≤ 109) — sizes of towers.

    Output

    Print the number of operations needed to destroy all towers.

    Examples
    Input
    6
    2 1 4 6 2 2
    Output
    3
    Input
    7
    3 3 3 1 3 3 3
    Output
    2
    Note

    The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color.

    After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation

    题目大意 每次只能消最外层的砖 问多少次能消完

    看hint图吧 等价与从右方看 从左到右峰依次为  2 1 4 3 2 1 从左方看  从左到右峰依次为 1 1 2 3 2 2

    比较每个位置需要消去的最少次数 依次为 1 1 2 3 2 2的峰

    看到这里是不是就明白了呢

    我们只需要把山峰等效为 突起 如 1 2 2 3 4 3这样的形式就ok了

    具体操作见代码 tw菊苣的dp写法还不是很理解 再研究一下

    #include<cstdio>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    inline void ri(int &num){
        num=0;int f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9')num=num*10+ch-'0',ch=getchar();
        num*=f;
    }
    const int N=1e5+5;
    int l[N],r[N],a[N];
    int main()
    {
        int n;
        ri(n);
        for(int i=1;i<=n;i++)ri(a[i]);
        for(int i=1;i<=n;i++) l[i]=min(l[i-1]+1,a[i]);
        for(int i=n;i>=1;i--) r[i]=min(r[i+1]+1,a[i]);
        int mx=-1;
        for(int i=1;i<=n;i++) mx=max(mx,min(l[i],r[i]));
        printf("%d
    ",mx);
        return 0;
    }
    AC代码
  • 相关阅读:
    《.NET分布式应用程序开》读书笔记 第一章:理解分布式架构
    一个DataSet的工具类,可以将DataTime的Time部分去掉,主要在序列化Xml时有用.
    Microsoft SQL Server 2005技术内幕系列书籍
    COM+客户端部署发现
    PowerDesigner中三种模型的转换关系图
    将ASP.NET页面内容输出到字符串中
    在WinForms中隐藏Crystal Report的[MainReport]标签页
    qmake常用语法一
    MinGW简介
    Qt的prx文件
  • 原文地址:https://www.cnblogs.com/Geek-xiyang/p/5402790.html
Copyright © 2011-2022 走看看