poj 2559 Largest Rectangle in a Histogram

Largest Rectangle in a Histogram

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24218		Accepted: 7830

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 3
4 1000 1000 1000 1000
0

Sample Output

8
4000

Hint

Huge input, scanf is recommended.

对于每一个H[i]，都可以找到一个L[i]  R[i]使得L[i]=(j<=i并且h[j-1]<h[i]的最大的j)

R[i]=(j>i并且h[j]>h[i]的最小的j)

H=min（h[i]|L<=i<R）

h[L-1]肯定是小于H的

h[R]肯定也是小于H的

 1° 用数组模拟栈

#include <cstdio>
#include <stack>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
	{
		int n,L[100005],R[100005];
		int h[100005];
		int st[100005];
		while (scanf("%d",&n)!=EOF&&n!=0)
		{
			//memset(h,0,sizeof(h));
			//memset(L,0,sizeof(L));
			//memset(R,0,sizeof(R));
			for (int i=0;i<n;i++)
				scanf("%d",&h[i]);
			int t=0;
			for (int i=0;i<n;i++)
			{
				while (t>0&&h[st[t-1]]>=h[i])//st 存储的是j<=i并且h[j-1]<h[i]最大的j
					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;
			}
			long long ans=0;
			for (int i=0;i<n;i++)
			ans=max(ans,(long long)h[i]*(R[i]-L[i]));
			printf("%lld\n",ans);
		}
	}

2° 使用栈解决

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <stack>
using namespace std;
int main()
    {
        int n,L[100005],R[100005];
        int h[100005];
        while (scanf("%d",&n)!=EOF&&n)
            {
                for (int i=0;i<n;i++)
                    scanf("%d",&h[i]);
                stack<int> s;
                //s.push(0);
                while (!s.empty()) s.pop();
                for (int i=0;i<n;i++)
                    {
                        while (!s.empty()&&h[s.top()]>=h[i])
                            s.pop();
                        if(s.empty()) L[i]=0;
                        else L[i]=s.top()+1;//因为s.top()是上一个位置，当前位置要加1
                        s.push(i);
                    }
                while (!s.empty()) s.pop();
                for (int i=n-1;i>=0;i--)
                    {
                        while (!s.empty()&&h[s.top()]>=h[i])
                            s.pop();
                        if(s.empty()) R[i]=n;
                        else R[i]=s.top();
                        s.push(i);
                    }
                long long ans=0;
                for (int i=0;i<n;i++)
                    ans=max(ans,(long long)h[i]*(R[i]-L[i]));
                    printf("%lld\n",ans);
            }
        return 0;
    }