hdu1506-动态规划

Largest Rectangle in a Histogram

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6051    Accepted Submission(s): 1737

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

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

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<map>#include<iomanip>#define INF 99999999using namespace std;const int MAX=100001;int Left[MAX],Right[MAX];//求和第i个木板连续的高于或等于它的最左边木板和最右边木板,i,j,然后面积s=(j-i+1)*s[i],求最大的s即可. __int64 s[MAX];int main(){int n;while(cin>>n,n){for(int i=0;i<n;++i){scanf("%I64d",&s[i]);Left[i]=Right[i]=i;//初始化连续的高于或等于它的木板只有它本身.}for(int i=1;i<n;++i){//求和第i个木板连续的高于或等于它的最左边的木板. while(Left[i]-1>=0 && s[Left[i]-1]>=s[i])Left[i]=Left[Left[i]-1];}for(int i=n-2;i>=0;--i){//求和第i个木板连续的高于或等于它的最右边的木板.while(Right[i]+1<n && s[Right[i]+1]>=s[i])Right[i]=Right[Right[i]+1];}__int64 sum=0;for(int i=0;i<n;++i){if(s[i]*(Right[i]-Left[i]+1)>sum)sum=s[i]*(Right[i]-Left[i]+1);}cout<<sum<<endl;}return 0;}