zoukankan      html  css  js  c++  java
  • 破解2559

    Description

    A histogram is a polygon composed of a sequence of rectangles aligned at a common base line. The rectangles have equal widths but may have different heights. For example, the figure on the left shows the histogram that consists of rectangles with the heights 2, 1, 4, 5, 1, 3, 3, measured in units where 1 is the width of the rectangles:       

    Usually, histograms are used to represent discrete distributions, e.g., the frequencies of characters in texts. Note that the order of the rectangles, i.e., their heights, is important. Calculate the area of the largest rectangle in a histogram that is aligned at the common base line, too. The figure on the right shows the largest aligned rectangle for the depicted histogram.      

    Input

    The input contains several test cases. Each test case describes a histogram and starts with an integer n, denoting the number of rectangles it is composed of. You may assume that 1<=n<=100000. Then follow n integers h1,...,hn, where 0<=hi<=1000000000. These numbers denote the heights of the rectangles of the histogram in left-to-right order. The width of each rectangle is 1. A zero follows the input for the last test case.     

    Output

    For each test case output on a single line the area of the largest rectangle in the specified histogram. Remember that this rectangle must be aligned at the common base line.      

    Sample Input

    7 2 1 4 5 1 3 3
    4 1000 1000 1000 1000
    0
    

    Sample Output

    8
    4000
    

    Hint

    Huge input, scanf is recommended.

    题意:给你一系列的矩形。宽度都为1。高度为h要求出相邻矩形连成的最大矩形面积。

    思路:

    假设确定了长方形的左端点L和右端点R,那么最大可能的高度就是min{hi|L <= i < R}。

    代码:

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int const maxn=100010;
    long long  h[maxn],left[maxn],right[maxn];
    void solve()
    {
        int n,temp;
        while(scanf("%d",&n)==1&&n)
        {
            memset(h,0,sizeof(h));
            memset(left,0,sizeof(left));
            memset(right,0,sizeof(right));
             for(int i=1;i<=n;i++)
                scanf("%I64d",&h[i]);
             for(int i=1;i<=n;i++)
                left[i]=right[i]=i;
                for(int i=2;i<=n;i++)
                {
                        temp=i;
                    while(h[temp-1]>=h[i]&&temp>1)
                            temp=left[temp-1];
                            left[i]=temp;
                }
                for(int i=n-1;i>0;i--)
                {
                    temp=i;
                    while(h[temp+1]>=h[i]&&temp<n)
                        temp=right[temp+1];
                        right[i]=temp;
    
                }
                long long  ans=0;
                for(int i=1;i<=n;i++)
                {
                    ans=max((long long )ans,h[i]*(right[i]-left[i]+1));
                }
                    printf("%I64d
    ",ans);
    
        }
    
    }
    int main()
    {
       solve();
        return 0;
    }
    


     


     

  • 相关阅读:
    c语言中程序的循环控制 大小值的判断及赋值
    python中猜数字小游戏
    R语言中自编函数(例题)
    c语言中continue语句
    c语言中程序的循环控制 变量的非常规变化例题
    python中向列表中添加元素
    mean
    python中原始字符串和长字符串
    ArcInfo 的工作空间和 Coverage
    ArcGIS资料大全
  • 原文地址:https://www.cnblogs.com/tlnshuju/p/6883176.html
Copyright © 2011-2022 走看看