破解2559

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.      

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

8
4000

Hint

Huge input, scanf is recommended.

题意：给你一系列的矩形。宽度都为1。高度为h要求出相邻矩形连成的最大矩形面积。

思路：

假设确定了长方形的左端点L和右端点R，那么最大可能的高度就是min{hi|L <= i < R}。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int const maxn=100010;
long long  h[maxn],left[maxn],right[maxn];
void solve()
{
    int n,temp;
    while(scanf("%d",&n)==1&&n)
    {
        memset(h,0,sizeof(h));
        memset(left,0,sizeof(left));
        memset(right,0,sizeof(right));
         for(int i=1;i<=n;i++)
            scanf("%I64d",&h[i]);
         for(int i=1;i<=n;i++)
            left[i]=right[i]=i;
            for(int i=2;i<=n;i++)
            {
                    temp=i;
                while(h[temp-1]>=h[i]&&temp>1)
                        temp=left[temp-1];
                        left[i]=temp;
            }
            for(int i=n-1;i>0;i--)
            {
                temp=i;
                while(h[temp+1]>=h[i]&&temp<n)
                    temp=right[temp+1];
                    right[i]=temp;

            }
            long long  ans=0;
            for(int i=1;i<=n;i++)
            {
                ans=max((long long )ans,h[i]*(right[i]-left[i]+1));
            }
                printf("%I64d\n",ans);

    }

}
int main()
{
   solve();
    return 0;
}