poj.2155.Matrix

发表于 2020-04-01 更新于 2020-08-05 分类于数据结构与算法， poj 阅读次数： Valine：

Matrix

问题描述

Description

Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N).

We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using “not” operation (if it is a ‘0’ then change it into ‘1’ otherwise change it into ‘0’). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions.

C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2).
Q x y (1 <= x, y <= n) querys A[x, y].
Input

The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case.

The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format “Q x y” or “C x1 y1 x2 y2”, which has been described above.
Output

For each querying output one line, which has an integer representing A[x, y].

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poj.2559.Largest Rectangle(单调栈)

发表于 2020-04-01 更新于 2020-04-09 分类于数据结构与算法， poj 阅读次数： Valine：

最大矩形

题目

2559.Largest Rectangle in a Histogram

描述

单调栈

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <vector>
#include <stack>
#include <queue>
#include <map>
#include <algorithm>
#include <ctime>
using namespace std;

long long LargetRectangle(int n, vector<int>& v)
{
	long long ans=0;
	stack<int> s;
	int temp = 0;
	int index = 0;
	long long cnt;

	for (int i = 0; i < n; i++)
	{
		while (!s.empty())
		{
			if (v[s.top()] > v[i])
			{
				index = s.top();
				s.pop();
				temp = s.empty() ? (i) : (i - s.top() - 1);
				cnt = (long long)v[index] * temp;
				ans = ans > cnt ? ans : cnt;
			}
			else
			{
				break;
			}
		}
		s.push(i);
	}

	while (!s.empty())
	{
		index = s.top();
		s.pop();
		temp = s.empty() ? (n) : (n - s.top() - 1);
		cnt = (long long)v[index] * temp;
		ans = ans > cnt ? ans : cnt;
	}

	return ans;
}
int main()
{
	int n;
	while (scanf("%d", &n)!= EOF)
	{
		if (n == 0)
		{
			break;
		}
		vector<int> v(n);
		for (int i = 0; i < n; i++)
		{
			scanf("%d", &v[i]);
		}
		printf("%lld\n", LargetRectangle(n, v));
	}

	return 0;
}