最大长方形（二）-OJ

最新推荐文章于 2024-03-28 08:00:00 发布

wang_fu_grong_16

最新推荐文章于 2024-03-28 08:00:00 发布

阅读量338

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/wang_fu_grong_16/article/details/77927870

本文探讨了如何在给定条件下找到二维数组中最大的矩形，这是计算机科学中常见的问题，涉及几何和动态规划思想。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

最大长方形（二）

时间限制：1000 ms | 内存限制：65535 KB

难度：4

描述

Largest Rectangle in a Histogram

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.

输入

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.

输出

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.

样例输入

7 2 1 4 5 1 3 3
4 1000 1000 1000 1000
0

样例输出

4000

个人理解：

结果	时间	内存	语言
Accepted	72	1800	c++

代码：

#include <stdio.h>
#include <string.h>
#include <stack>
using namespace std;

#define max(a,b)((a)>(b)?(a):(b))

long long int stack1[100000+10];
long long int len[100000+10];

int main()
{
int n;
while(~scanf("%d",&n),n)
{
int t=0;
stack1[t]=-2; //这个值与下面的那个-1，相照应，必须要小于下面的那个数，否则程序错误
long long int max1=0,h;
memset(len,0,sizeof(len));
for(int i=0;i<=n;i++)
{
if(i<n)
scanf("%lld",&h);
else h=-1;
if(h>stack1[t])
{
stack1[++t]=h;
len[t]=1;
}
else
{
long long int cont=0,k=0;
while(h <= stack1[t])
{
max1=max(max1,(k+len[t])*stack1[t]); //同步更新，寻找最大的面积。
k+=len[t--]; //t的减少，表示覆盖了那些双层循环中重复的部分
} //k的增加，表示了当前stack1[t] 前面有k个大于等于它的数
stack1[++t]=h; //len[t]值更新加上k,表示当前值h的前面有k个大于等于它的数
len[t]=k+1; //且这一步就是去掉重复部分的关键
}
}
printf("%lld\n",max1);
}
return 0;
}