uva 1506 Largest Rectangle in a Histogram

Largest Rectangle in a Histogram

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

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

 

Source

University
of Ulm Local Contest 2003

 

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题意：有n个数ai。从中选出一段区间[L,R]，使得(R-L+1)*min{a_L,…,a_R}最大。

经典广告印刷问题

考虑对于第i个数，求出当这个数成为最小值时，往左往右分别最远能到哪里。

使用单调队列来实现这一过程。

 

#include<cstdio>
#include<algorithm>
#define N 100001

#ifdef WIN32
#define ll "%I64d\n"
#else
#define ll "%lld\n"
#endif

using namespace std;
int n,a[N],b[N];
int q[N],tmp[N],head,tail;
int l[N],r[N];
long long ans;
void monotonous(int *c,int *d)
{
    q[]=c[]; tmp[]=;
    head=; tail=;
    for(int i=;i<=n;i++)
    {
        if(c[i]<q[tail-])
            while(head<tail && q[tail-]>c[i])    d[tmp[--tail]]=i-;
        q[tail]=c[i];
        tmp[tail++]=i;
    }
    while(head<tail) d[tmp[head++]]=n;
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        if(!n) return ;
        ans=;
        for(int i=;i<=n;i++) scanf("%d",&a[i]),b[n-i+]=a[i];
        monotonous(a,r);
        monotonous(b,l);
        for(int i=;i<=n;i++) tmp[i]=l[i];
        for(int i=;i<=n;i++) l[n-i+]=n-tmp[i]+;
        for(int i=;i<=n;i++) ans=max(ans,1ll*(r[i]-l[i]+)*a[i]);
        printf(ll,ans);
    }
}