单调栈poj2559
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Largest Rectangle in a Histogram
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 25276 | Accepted: 8167 |
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.
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.
n的范围是1e5，即使是用n*n的动态规划来写，也是会超时的。
我们用单调栈可以在O(n)的复杂度下解决。
单调栈，也就是栈内的元素是按照某种顺序有序排列的。
在新的元素将要入栈的时候，如果满足单调性，就直接入栈。
如果不满足的话，就弹出栈内的元素，直到满足后入栈。
用法：用来求出某个数的左边和右边第一个大于或小于它的元素，时间复杂度O(n)。
我们枚举每一个小矩形的高度，以该高度为最低限度，然后向两边进行延伸。
向左延伸的时候，延伸到第一个小于该最低限度的矩形的时候结束，向右也是，
延伸到第一个小于该最低限度的矩形的时候结束。
也就是求左边和右边第一个小于它的元素
因为我们可以用单调栈来解决。
维护一个严格单调递增的单调栈。
向左求出第一个小于它的元素的位置+1，也就是可以想左延伸的最长距离
向右求出第一个大于它的元素的位置-1，也就是可以向右延伸的最长距离
AC代码
#include
#include
#include
#define maxn 100005
#define LL long long
using namespace std;
int n;
LL h[maxn];
LL l[maxn];
LL r[maxn];
stack s;
int main()
{
while(~scanf("%d",&n))
{
if(n==0)
break;
for(int i=1; i<=n; i++)
{
scanf("%lld",&h[i]);
}
while(!s.empty())
{
s.pop();
}
for(int i=1; i<=n; i++)
{
while(!s.empty()&&h[s.top()]>=h[i])
{
s.pop();
}
if(s.empty())
l[i]=1;
else
l[i]=s.top()+1;
s.push(i);
}
while(!s.empty())
{
s.pop();
}
for(int i=n; i>=1; i--)
{
while(!s.empty()&&h[s.top()]>=h[i])
{
s.pop();
}
if(s.empty())
r[i]=n;
else
r[i]=s.top()-1;
s.push(i);
}
LL ans=-1;
for(int i=1; i<=n; i++)
{
ans=max(ans,h[i]*(r[i]-l[i]+1));
}
printf("%lld\n",ans);
}
return 0;
}