poj 2559 Largest Rectangle in a Histogram

佚名 6年前 (2019-04-25) 随笔 1817人围观抢沙发百度已收录

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:

poj 2559 Largest Rectangle in a Histogram 随笔

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

SRE实战互联网时代守护先锋，助力企业售后服务体系运筹帷幄！一键直达领取阿里云限量特价优惠。 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,...,h_n, where 0<=h_i<=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

Hint

Huge input, scanf is recommended.

Source

Ulm Local 2003 单调栈求左右边界。代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>

using namespace std;
typedef long long ll;
int n;
ll in[100002];
int l[100002],r[100002];
int main() {
    stack<int> s;
    while(~scanf("%d",&n) && n) {
        while(!s.empty()) s.pop();
        in[++ n] = -1;
        for(int i = 1;i <= n;i ++) {
            if(i < n) scanf("%lld",&in[i]);
            l[i] = r[i] = i;
            while(!s.empty() && in[s.top()] >= in[i]) {
                l[i] = l[s.top()];
                r[s.top()] = i - 1;
                s.pop();
            }
            s.push(i);
        }
        ll ans = 0;
        for(int i = 1;i < n;i ++) {
            ans = max(ans,in[i] * (r[i] - l[i] + 1));
        }
        printf("%lld\n",ans);
    }
}