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最新推荐文章于 2024-05-11 14:57:22 发布

转载最新推荐文章于 2024-05-11 14:57:22 发布 · 89 阅读

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原文链接：http://www.cnblogs.com/tonix/p/4988677.html

本文介绍了一种改进的Largest Rectangle算法实现方法，通过两次运行算法来解决特定问题，详细步骤包括输入处理、使用栈进行计算以及最终输出最大矩形面积。

Actually I think problem statement is somewhat misleading. No need to mention range [L, R] at all.

The intention is a variation to "Largest Rectangle" which is a classic stack problem on LeetCode.

But you need to run Largest Rectangle twice: increased and decreased.

#include <cmath>
#include <stack>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    int n; cin >> n;
    vector<int> in(n);
    for(int i = 0; i < n; i ++)
        cin >> in[i];
    
    int ret = 0;
    {
        in.push_back(0); 
        stack<int> stk;
        for (int i = 0; i < in.size(); i++)
        {
            if (stk.empty() || in[i] > stk.top())
            {
                stk.push(in[i]);
            }
            else
            {
                int lastV = stk.top(); stk.pop();
                if (!stk.empty())
                {
                    ret = std::max(ret, (lastV ^ stk.top()));            
                    //cout << lastV << "-" << stk.top() << "=" << (lastV ^ stk.top()) << endl;
                }
                i--;
            }
        }
        in.pop_back();
    }
    {
        
        stack<int> stk;
        in.insert(in.begin(), 0);
        for (int i = in.size() - 1; i >= 0; i --)
        {
            if (stk.empty() || in[i] > stk.top())
            {
                stk.push(in[i]);
            }
            else
            {
                int lastV = stk.top(); stk.pop();
                if (!stk.empty())
                {
                    ret = std::max(ret, (lastV ^ stk.top()));            
                    //cout << lastV << "-" << stk.top() << "=" << (lastV ^ stk.top()) << endl;
                }
                i++;
            }
        }
    }
    
    cout << ret << endl;
    return 0;
}