Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20760    Accepted Submission(s): 6325

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  

 

Recommend LL   |   We have carefully selected several similar problems for you:  1505 1069 1087 1058 1176    一道挺裸的单调栈 我们预处理出每一个位置向左/向右第一个比他小的位置 然后统计答案，这样很显然是对的 时间复杂度$O(N)$  

#include<cstdio>
#define LL long long 
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
using namespace std;
const int MAXN = 100001;
LL a[MAXN];
int s[MAXN], top = 0;
int L[MAXN], R[MAXN];
void clear() {top = 0;}
int main() {
    //#ifdef WIN32
    //freopen("a.in", "r", stdin);
    //#endif
    int N;
    while(~scanf("%d", &N) && N != 0) {
        clear();
        s[0] = 0;
        for(int i = 1; i <= N; i++) scanf("%I64d",&a[i]);
        for(int i = 1; i <= N; i++) {
            while(top > 0 && a[i] <= a[ s[top] ]) top--;
            L[i] = min(s[top] + 1, i);
            s[++top] = i;
        }
        clear();
        s[0] = N + 1;
        for(int i = N; i >= 1; i--) {
            while(top > 0 && a[i] <= a[ s[top] ]) top--;
            R[i] = max(s[top] - 1, i);
            s[++top] = i;
        }
        LL Ans = 0;
        for(int i = 1; i <= N; i++) 
            Ans = max(Ans, (R[i] - L[i] + 1) * a[i] );
        printf("%I64d\n",Ans);
    }
    return 0;
}