题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1506

Problem 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

题目大意：给一个直方图，然后找到能够构成一个矩形的面积最大的连续的方块，两张图片已经给出来了什么是矩形面积的最大的联通块。

对于每个柱子，找到能够到达的左边最远的地方和右边最远的地方，然后找出最大的面积就行了

ac：


#include<stdio.h>  
#include<string.h>  
#include<math.h>  
  
//#include<map>   
#include<deque>  
#include<queue>  
#include<stack>  
#include<string>  
#include<iostream>  
#include<algorithm>  
using namespace std;  
  
#define ll long long  
#define da    0x3f3f3f3f  
#define xiao -0x3f3f3f3f  
#define clean(a,b) memset(a,b,sizeof(a))// 雷打不动的头文件  

ll Left[100100],Right[100100];
ll shuzu[100100];

int main()
{
	int n;
	while(cin>>n)
	{
		clean(Left,0);
		clean(Right,0);
		clean(shuzu,0);
		if(n==0)
			break;
		int i,j;
		for(i=1;i<=n;++i)
			scanf("%lld",&shuzu[i]);
		//找左边的边界 
		Left[1]=1;//最左边是第一块 
		for(i=2;i<=n;++i)//从第二个开始 
		{
			j=i; //向左查找，如果该值>=当前值，就记录位置并且继续 
			while(j>1&&shuzu[j-1]>=shuzu[i])
				j=Left[j-1];//这里Left中储存的是 第几块 
			Left[i]=j;//最后找到的是能到达的 最左边的距离 
		}
		//同理 右边的边界 
		Right[n]=n;//最右边是第n块 
		for(i=n-1;i>=1;--i)//从n-1开始 
		{
			j=i;//向右查找，如果该值>=当前值，继续 
			while(j<n&&shuzu[j+1]>=shuzu[i])
				j=Right[j+1];//right中是最右边能到达的地方 
			Right[i]=j;//最后找到最右边的距离 
		}
		ll maxx=0;
		for(i=1;i<=n;++i)//遍历；找 所有面积并找最大的 
			maxx=maxx>(Right[i]-Left[i]+1)*shuzu[i]?maxx:(Right[i]-Left[i]+1)*shuzu[i];
		cout<<maxx<<endl;//输出maxx 
	}
}