題目連結：http://poj.org/problem?id=2559

Largest Rectangle in a Histogram

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 20199	Accepted: 6489

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 integersh₁,...,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.

尷尬，只會n^2做法，只能水過學校OJ的資料，水不過poj原題,.........

沒想到就是沒想到，菜得摳腳.......

網上有個比較巧的思路:

使用一個叫“單調棧”的東西

想法很巧妙：將木板從小到大扔進棧中，一旦手裡拿的準備進棧的木板（稱之為A板）比棧頂的木板還大

就把棧頂的木板拉出來（稱之為B板），統計一下面積，與ans比較，存起來最大的，

然後再把棧頂的板拉出來（稱之為C板），如果還是小於準備進棧的A板，那麼把B板切成C板大小，

統計“兩塊C板”的面積，與ans比較，存最大（原因：因為棧是單調棧，從小到大存放，那麼拉出來的C板肯定比B板小

那麼B板肯定可以切成C板大小，那麼統計時就有兩塊C板合成的大板），如此往復......

直到棧頂的板終於不大於A板了，把之前那一堆疊頂彈出，不斷切小的板（B板C板等的疊起來的n塊板）

切成A板大小,與A板釘在一起扔進棧中，那麼此時相當於有 n + 1 塊A板合起來的大板（理解為面積）

切成和手裡的板的大小相同，在將手裡的兩塊板疊成一塊的（理解為面積不變，厚度增加......）

最後把所以棧中的存板全扔出來切，看最多能切成面積多大的板.......

為何能這麼幹？？

首先：單調棧控制棧中板大小遞增

其次：一旦手裡板小於棧頂板，這意味著沒有其他比棧頂板更大的板可以切成棧頂的板，那麼面積為棧頂板大小的板只有1塊
（這裡還有一個前提：“合成板”只能是自身前後相鄰的幾塊組成，不能跳過某塊板與其他“合成”）

#include <iostream>
#include <algorithm>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <queue>
#include <cstdio>
#include <string.h>
#include <cstdlib>
#include <cmath>
#include <string>
typedef long long ll;
#define CLR(a) memset(a, 0, sizeof(a))
#define SD(a) scanf("%d", &a)
#define FOR(i, a, b)  for(i = a; i < b; i++)
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define MAXN 100000 + 10
using namespace std;

typedef struct
{
     ll width;
     ll height;
}Block;

Block arr[MAXN];
int main()
{
     int n;
     while (~scanf("%d", &n) && n > 0)
     {
          ll ans = 0;
          
          stack<Block> blockStack;
          for (int i = 0; i < n; i++)
          {
               scanf("%lld", &arr[i].height);
               ll tmp = 0;
               while (blockStack.empty() == false && arr[i].height < blockStack.top().height)
               {
                    ans = max(ans, (blockStack.top().width + tmp) * blockStack.top().height);
                    tmp += blockStack.top().width;
                    blockStack.pop();
               }
               arr[i].width = tmp + 1;
               blockStack.push(arr[i]);
          }

          ll tmp = 0;
          while (blockStack.empty() == false)
          {
               ans = max(ans, (blockStack.top().width + tmp) * blockStack.top().height);
               tmp += blockStack.top().width;
               blockStack.pop();
          }
          printf("%lld\n", ans);
     }

     return 0;
}