HDU1506 Largest Rectangle in a Histogram(水DP)

Largest Rectangle in a Histogram

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16166 Accepted Submission(s): 4747

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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注意一下结果会超过int用long long存，维护l和r数组，l[i]存储的是i位置方块左边最后一个高度大于等于它的方块位置，同理r存的是右边的位置。用dp维护一下就阔以了。

#include "cstring"
#include "cstdio"
#include "iostream"
#include "string.h"
using namespace std;
long long a[100005],l[100005],r[100005];
int main()
{
    int n;
    while(scanf("%d",&n)&&n!=0)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
            l[i]=i;
            r[i]=i;
        }
        memset(l,0,sizeof(l));
        memset(r,0,sizeof(r));
        l[1]=1;
        r[n]=n;
        for(int i=2;i<=n;i++)
        {
            long long t=i;
            while(t>1&&a[i]<=a[t-1])
                t=l[t-1];
            l[i]=t;
        }
        for(int i=n;i>=1;i--)
        {
            long long t=i;
            while(t<n&&a[i]<=a[t+1])
                t=r[t+1];
            r[i]=t;
        }
        long long ans=0;
        for(int i=1;i<=n;i++)
        {
            long long now;
            now=(r[i]-l[i]+1)*a[i];
            if(ans<now)
                ans=now;
        }
        printf("%lld\n",ans);
    }

}