Largest Rectangle in a Histogram

 题目传送门：http://poj.org/problem?id=2559

 

Largest Rectangle in a Histogram

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24256		Accepted: 7841

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 3
4 1000 1000 1000 1000
0

Sample Output

8
4000

Hint

Huge input, scanf is recommended.

Source

Ulm Local 2003

 

　题解：

　　单调栈计算每个矩形往左、往右所能够达到的最远距离。（入栈判断左边界，出栈判断右边界）

　　PS：我用的是左开右闭区间。

　　一个小技巧：栈中直接维护位置信息，这样就不用开两个栈或是结构体了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stack>
#define LL long long
#define RI register int
using namespace std;
const int INF = 0x7ffffff ;
const int N = 100000 + 10 ;

inline LL read() {
    int k = 0 , f = 1 ; char c = getchar() ;
    for( ; !isdigit(c) ; c = getchar())
      if(c == '-') f = -1 ;
    for( ; isdigit(c) ; c = getchar())
      k = k*10 + c-'0' ;
    return k*f ;
}
LL n ; LL hh[N], l[N] ;
stack<LL>s ;  // 直接存位置信息 

inline void solve() {
//    while(s.size()) s.pop() ;
    LL ans = 0 ;
    for(int i=1;i<=n;i++) hh[i] = read() ;
    for(int i=1;i<=n;i++) {
        while(s.size() && hh[s.top()] >= hh[i]) {
            ans = max(ans,hh[s.top()]*(i-l[s.top()]-1)) ; s.pop() ;
        }
        if(s.size()) l[i] = s.top() ; else l[i] = 0 ;  // 初始化l数组的话这里就不用加else了
        s.push(i) ;  // printf("%d %d
",i,l[i]) ;
    }
    while(s.size()) ans = max(ans,(n-l[s.top()])*hh[s.top()]), s.pop() ;
    printf("%lld
",ans) ;
}

int main() {
    while(1) {
        n = read() ; if(!n) return 0 ;
        solve() ;
    }
}