POJ - 2559（单调栈入门题）

Largest Rectangle in a Histogram

Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

Submit Status

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 34 1000 1000 1000 10000

Sample Output

84000

题意：给你n个连续的矩形，他们的宽度都是1，高度不同。连接起来形成一个不规则的图像。问这个图像中最大的矩形面积是多少？

题解：使用栈去维护这个高度，如果当前的高度h大于栈顶就入栈，否则一直弹出到比这个数字小为止，因为这个栈一直是单调递增的，那么我们称这样的栈是单调栈。这一题也是单调栈的入门题

#include<iostream>#include<cstdio>using namespace std;#define N 100005 #define LL long longint n;LL h[N],L[N],R[N],st[N];void solve(){int t=0;for(int i=0;i<n;i++){while(t>0&&h[st[t-1]]>=h[i])t--;L[i]=t==0?0:(st[t-1]+1);st[t++]=i;}t=0;for(int i=n-1;i>=0;i--){while(t>0&&h[st[t-1]]>=h[i])t--;R[i]=t==0?n:st[t-1];st[t++]=i;}LL res=0;for(int i=0;i<n;i++){res=max(res,(h[i]*(R[i]-L[i])));}printf("%lld\n",res);}int main(){#ifdef CDZSCfreopen("i.txt","r",stdin);#endifwhile(~scanf("%d",&n)){if(n<=0)break;for(int i=0;i<n;i++)scanf("%d",h+i);solve();}return 0;}