HDU 1506 Largest Rectangle in a Histogram

Largest Rectangle in a Histogram

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 1506
64-bit integer IO format: %I64d      Java class name: Main

 

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

 

解题：单调队列的维护。。。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 100010;
int q[maxn] = {-1},w[maxn],tp,n,h;//w记录前面比自己大的元素个数
LL ans;
int main(){
    while(scanf("%d",&n),n){
        ans = tp = 0;
        for(int i = 0; i <= n; i++){
            if(i == n) h = 0;
            else scanf("%d",&h);
            if(h > q[tp]){
                q[++tp] = h;
                w[tp] = 1;
            }else{
                LL cnt = 0;
                while(h <= q[tp]){
                    ans = max(ans,(cnt+w[tp])*q[tp]);//每次找q中下降的
                    cnt += w[tp--];
                }
                q[++tp] = h;
                w[tp] = ++cnt;//算上自己
            }
        }
        printf("%I64d
",ans);
    }
    return 0;
}