POJ 2559 单调栈

Largest Rectangle in a Histogram

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 17222 Accepted: 5562

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,...,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 34 1000 1000 1000 10000

Sample Output

84000

Hint

Huge input, scanf is recommended.

单调栈在好久以前就写过了～但是突然看到这个裸的单调栈的题目居然有点茫然

于是决定写个博客来记录一下～

思路就是用单调栈找出每个点能向左向右的贡献度，然后暴力扫描一遍即可～

#include <iostream>#include <cstdio>#include <cmath>#include <string>#include <cstring>#include <algorithm>#include <queue>#include <map>#include <set>#include <stack>#include <sstream>#define PI acos(-1.0)const int  inf =  (1<<30) - 10;using namespace std;const int maxx = 100000 + 10;long long num[maxx],L[maxx],R[maxx];long long ans;int n;stack<long long> p;int main(){    while(cin>>n && n){        for(int i = 1;i <= n; ++i){            scanf("%lld",&num[i]);        }        p.push(-1);        for(int i = 1;i <= n; ++i){            while(p.top() != -1 && num[p.top()]>num[i]){                R[p.top()] = i - p.top() - 1;                p.pop();            }            p.push(i);        }        while(p.top()!=-1){            R[p.top()] = n - p.top();            p.pop();        }        for(int i = n; i >= 1; --i){            while(p.top() != -1 && num[p.top()] > num[i]){                L[p.top()] = p.top() - i;                p.pop();            }            p.push(i);        }        while(p.top() != -1){            L[p.top()] = p.top() - 0;            p.pop();        }        ans = 0;        for(int i = 1;i <= n; ++i){            ans = max(ans,num[i]*(L[i]+R[i]));        }        printf("%lld\n",ans);    }    return 0;}

如有BUG，欢迎指出～