hdu 1506 Largest Rectangle in a Histogram

Largest Rectangle in a Histogram

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9075    Accepted Submission(s): 2501

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 34 1000 1000 1000 10000

 

Sample Output

84000

 

Source

University of Ulm Local Contest 2003

 

Recommend

LL

 

dp。

思路：分别用两个数组记录对应的每个矩形左侧和右侧高度大于等于它本身的矩形的数量，这样用(高度*(左侧数量+右侧数量+1))即可表示对应高度的最大矩形面积。

100000的数据量用O(n2)的暴力肯定会T,分别从左到右更新左侧数量和从右到左更新右侧数量。刚开始用long long 一直wa,后来改成__int64就过了。

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<map>#include<queue>#include<stack>#include<vector>#include<ctype.h>#include<algorithm>#include<string>#define PI acos(-1.0)#define maxn 100005#define INF 1<<25__int64 hh[maxn];__int64 le[maxn];__int64 ri[maxn];using namespace std;int main(){    __int64 tot;    while(scanf("%I64d",&tot))    {        if(tot==0)        break;        memset(hh,0,sizeof(hh));        memset(le,0,sizeof(le));        memset(ri,0,sizeof(ri));        le[0]=0;        ri[tot-1]=0;        ll i,j;        for(i=0;i<tot;i++)        {            scanf("%I64d",&hh[i]);            j=i-1;            while(hh[j]>=hh[i]&&j>=0)            {                j-=(le[j]+1);            }            le[i]=i-j-1;        }        for(i=tot-1;i>=0;i--)        {            j=i+1;            while(hh[j]>=hh[i]&&j<=tot-1)            {                j+=(ri[j]+1);            }            ri[i]=j-i-1;        }        __int64 ans=0;        for(i=0;i<tot;i++)        {            if((ri[i]+le[i]+1)*hh[i]>ans)            ans=(ri[i]+le[i]+1)*hh[i];        }        printf("%I64d\n",ans);    }}