HDU - 1506 Largest Rectangle in a Histogram(最大子矩阵)

最新推荐文章于 2019-07-31 17:44:00 发布

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最新推荐文章于 2019-07-31 17:44:00 发布

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分类专栏：题库_HDU DP ACM 文章标签： dp hdu

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本文介绍了一种高效算法，用于解决给定一组矩形高度后寻找最大矩形面积的问题。通过两次遍历实现，首先从左至右确定左侧较高边界，接着从右至左确定右侧较高边界。

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点击打开题目链接

Largest Rectangle in a Histogram

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19253 Accepted Submission(s): 5803

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 3 4 1000 1000 1000 1000 0

Sample Output

8 4000

Source

University of Ulm Local Contest 2003

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题目大意：

如图，对于给出的N个矩阵，求出最大的连续面积。

思路：

对于每个矩阵h[i]，找出左边高度连续大于等于它的最远矩阵位置l，找出右边高度连续大于等于它的最远矩阵位置r。则面积为(r-l+1)*h[i];

问题则成为求l和r数组。从左到右遍历一遍求l[],再从右往左遍历一遍求r[]。

附上AC代码：

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=100000+5;
typedef long long ll;
ll h[maxn];
int l[maxn],r[maxn];
int n,t;
ll ans;

int main(){
    ios::sync_with_stdio(false);
    while(cin>>n,n){
        for(int i=0;i<n;i++){
            cin>>h[i];
        }
        l[0]=0;
        r[n-1]=n-1;
        for(int i=1;i<n;i++){
            t=i;
            while(t>0&&h[t-1]>=h[i])
                t=l[t-1];
            l[i]=t;
        }
        for(int i=n-2;i>=0;i--){
            t=i;
            while(t<n-1&&h[t+1]>=h[i])
                t=r[t+1];
            r[i]=t;
        }
        ans=0;
        for(int i=0;i<n;i++){
            if((r[i]-l[i]+1)*h[i]>ans)
                ans=(r[i]-l[i]+1)*h[i];
        }
        cout<<ans<<endl;
    }
    return 0;
}