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2013
11-11

POJ 2559 Largest Rectangle in a Histogram [解题报告] Java

Largest Rectangle in a Histogram

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

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

样例输出:

8
4000

温馨提示:

Huge input, scanf is recommended.

解题代码:

//* @author: [email protected]
import java.io.*;
public class Main
{
 static int[] p,b,c;
 static int a;
 public static void main(String[] args) throws IOException
 {
  InputStreamReader is=new InputStreamReader(System.in);
  BufferedReader in=new BufferedReader(is);
  while(true)
  {
   String[] ss=in.readLine().split(" ");
   a=Integer.parseInt(ss[0]);
   if(a==0) break;
   p=new int[a+2];
   b=new int[a+1];
   c=new int[a+1];
   for(int i=1;i<=a;i++)
	p[i]=Integer.parseInt(ss[i]);
   p[0]=p[a+1]=-1;
   long max=0;
   for(int i=1;i<=a;i++)
   {
	b[i]=1;
	int j=i;
	while(p[j-b[j]]>=p[i])
	{
          j-=b[j];
	  b[i]+=b[j];
	}
    }
    for(int i=a;i>0;i--)
    {
	c[i]=1;
	int j=i;
	while(p[j+c[j]]>=p[i])
	{
	  j+=c[j];
	  c[i]+=c[j];
	}
     }
    for(int i=1;i<=a;i++)
     {
	long h=(long)p[i]*(b[i]+c[i]-1);
	max=Math.max(h, max);
     }
    System.out.println(max);
  }
 }
}