感覺動態規劃都是玄妙的很，思維題吧（單調棧思維）

題解：讓求最大矩形面積，寬為1，暴力超時

可以發現 ? 當第i-1個比第i個高的時候 ? 比第i-1個高的所有也一定比第i個高 ?

于是可以用到動態規劃的思想

令left[i]表示包括i在內比i高的連續序列中最左邊一個的編號 ??right[i]?為最右邊一個的編號

那么有 ? 當?h[left[i]-1]>=h[i]]時 ? left[i]=left[left[i]-1]??從前往后可以遞推出left[i] ??

同理 ? ? ?當?h[right[i]+1]>=h[i]]時 ? right[i]=right[right[i]+1]?? 從后往前可遞推出righ[i]

最后答案就等于??max((right[i]-left[i]+1)*h[i])?了；

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 /*題目題意:題目給了n個矩形的高度，問最大連續矩形的公共面積（底乘以這段連續矩形中 最短的高度)，每個矩形的底是1 題目分析:我們可以枚舉每一個矩形，把它當作最矮的矩形，剩下就差知道底了。 既然這個矩形是最矮的的那一個，那么它左邊的矩形和右邊的矩形的高度應該大于等于它！*/ #include<iostream> #include<stdio.h> #include<cstring> #include<cmath> #define ll long long using namespace std; const int maxn=1e5+1000; ll a[maxn],Left[maxn],Right[maxn]; int main() {int n;while (~scanf("%d",&n)&&n){for(int i=1; i<=n; i++)scanf("%lld",&a[i]);Left[1]=1;Right[n]=n;for (int i=2; i<=n; i++) ///求出每個矩形左端非遞減連續的下標{int t=i;while (t>1&&a[i]<=a[t-1])/**狀態轉移*/t=Left[t-1];Left[i]=t;}for (int i=n-1; i>=1; i--) ///求出每個矩形右端非遞減連續的下標{int t=i;while (t<n&&a[i]<=a[t+1])t=Right[t+1];Right[i]=t;}ll ans=0;for (int i=1; i<=n; i++)ans=max(ans,(Right[i]-Left[i]+1)*a[i]);printf("%lld\n",ans);}return 0; }

?

總結

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