HDU 1506 Largest Rectangle in a Histogram(区间DP)

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

题目：

Largest Rectangle in a Histogram

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18833    Accepted Submission(s): 5636

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 usedto 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

 

思路：

循环遍历每个小矩形，找到它左边连续的不小于他高度的矩形下标left[i]，找到它右边连续的不小于他高度的矩形下标right[i]。

则若干小矩形形成的大矩形宽为right[i]-letf[i]+1,面积就等于(right[i]-left[i]+1)*a[i]，a[i]为当前遍历的到的小矩形高度。

结果：res=max(res,(right[i]-left[i]+1)*a[i])。

 

代码：

 #include <cstdio>
 #include <algorithm>
 using namespace std;
 typedef long long ll;
 const int N=;
 int n,a[N];
 int left[N],right[N];
 int main(){
     while (scanf("%d",&n)!=EOF && n) {
         ll res=;
         for (int i=; i<=n; i++) {
             scanf("%d",&a[i]);
             left[i]=i;
             right[i]=i;
         }
         for (int i=; i<=n; i++) {
             int j=i;
             while (j>= && a[i]<=a[j])  j=left[j]-;//通过之前的遍历结果更快地找到区间，若是用j--;会超时
             left[i]=j+;
         }
         for (int i=n-; i>=; i--) {
             int j=i;
             while (j<=n && a[i]<=a[j])  j=right[j]+;
             right[i]=j-;
         }
         for (int i=; i<=n; i++)    res=max(res,1LL*(right[i]-left[i]+)*a[i]);//转换为long long型
         printf("%lld\n",res);
     }
     return ;
 }