HDU-1506 Largest Rectangle in a Histogram (线性dp 维护前后边界)

Largest Rectangle in a Histogram

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19775    Accepted Submission(s): 5994

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

#include <bits/stdc++.h>using namespace std;int a[100001], l[100001], r[100001];int main(){int n;while(scanf("%d", &n) != EOF){if(n == 0) return 0;for(int i = 1; i <= n; ++i){scanf("%d", &a[i]);}for(int i = 1; i <= n; ++i){l[i] = r[i] = i;}a[0] = -1;for(int i = 2; i <= n; ++i){while(a[i] <= a[l[i] - 1]){l[i] = l[l[i] - 1];}}a[n + 1] = -1;for(int i = n - 1; i >= 1; --i){while(a[i] <= a[r[i] + 1]){r[i] = r[r[i] + 1];}}long long ans = 0;for(int i = 1; i <= n; ++i){ans = max(ans, 1LL * (r[i] - l[i] + 1) * a[i]);}printf("%lld\n", ans);}}/*题意：1e5的点，每个点有一个值，表示这里有一条垂直与X的边，两条边之间上端点可以连接当且仅当两边之间没有更短的边，问用这些边围成的矩形面积最大为多少。思路：对于每条边，记录左边最左一条比自己短的边，右边最右一条比自己短的边。然后高为该边的矩形即为a[i] * (r - l + 1)。枚举所有点即可维护答案即可。*/