zoukankan      html  css  js  c++  java
  • POJ 2559 Largest Rectangle in a Histogram

    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 18942   Accepted: 6083

    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 follown 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.

    Source

    刚开始竟然没有注意到区间长度的累计,各种算错。

    维护一个单增的单调栈,每步更新区间长度和答案即可。

     1 /*by SilverN*/
     2 #include<iostream>
     3 #include<algorithm>
     4 #include<cstring>
     5 #include<cstdio>
     6 #include<cmath>
     7 using namespace std;
     8 const int mxn=100020;
     9 struct stk{
    10     long long h,p;
    11 }st[mxn];
    12 int top;
    13 int n;
    14 long long ans;
    15 int main(){
    16     while(scanf("%d",&n) && n){
    17         ans=0;
    18         int i,j;
    19         int num;
    20         for(i=1;i<=n+1;i++){
    21             if(i>n)num=0;//数据读完后清栈 
    22             else scanf("%I64d",&num);
    23             if(num>=st[top].h) st[++top].h=num,st[top].p=1;//计算新区间 
    24             else{
    25                 int len=0;
    26                 while(top && num<=st[top].h){
    27                     len+=st[top].p;
    28                     ans=max(ans,len*st[top].h);
    29                     top--;
    30                 }
    31                 st[++top].p=len+1;//累计长度 
    32                 st[top].h=num;//更新高度 
    33             }
    34         }
    35         printf("%I64d
    ",ans);
    36     }
    37     return 0;
    38 }
  • 相关阅读:
    级联操作
    深入解析http协议
    http状态码
    数据库在一对一、一对多、多对多怎么设计表关系
    [转载]C#深拷贝的方法
    TraceSource记录程序日志
    .NET 垃圾回收与内存泄漏
    DevExpress GridControl使用方法总结
    DevExpress使用技巧总结
    Oracle 错误代码小结
  • 原文地址:https://www.cnblogs.com/SilverNebula/p/5651644.html
Copyright © 2011-2022 走看看