777C Alyona and Spreadsheet

最新推荐文章于 2020-11-12 23:38:10 发布

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本文介绍了一个关于表格排序的问题，即给定一个整数表格，在指定的行范围内判断是否存在至少一列是非递减排列的。通过动态规划的方法进行优化，使得查询效率大大提高。

During the lesson small girl Alyona works with one famous spreadsheet computer program and learns how to edit tables.

Now she has a table filled with integers. The table consists of n rows and m columns. By a_i, j we will denote the integer located at the i-th row and the j-th column. We say that the table is sorted in non-decreasing order in the column j if a_i, j ≤ a_{i + 1, j} for all i from 1 to n - 1.

Teacher gave Alyona k tasks. For each of the tasks two integers l and r are given and Alyona has to answer the following question: if one keeps the rows from l to r inclusive and deletes all others, will the table be sorted in non-decreasing order in at least one column? Formally, does there exist such j that a_i, j ≤ a_{i + 1, j} for all i from l to r - 1 inclusive.

Alyona is too small to deal with this task and asks you to help!

Input

The first line of the input contains two positive integers n and m (1 ≤ n·m ≤ 100 000) — the number of rows and the number of columns in the table respectively. Note that your are given a constraint that bound the product of these two integers, i.e. the number of elements in the table.

Each of the following n lines contains m integers. The j-th integers in the i of these lines stands for a_i, j (1 ≤ a_i, j ≤ 10⁹).

The next line of the input contains an integer k (1 ≤ k ≤ 100 000) — the number of task that teacher gave to Alyona.

The i-th of the next k lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n).

Output

Print "Yes" to the i-th line of the output if the table consisting of rows from l_i to r_i inclusive is sorted in non-decreasing order in at least one column. Otherwise, print "No".

输入一个矩阵A，取矩阵的第l行至第r行，问其中是否有一列是非递减排列的。

动态规划。首先想到设dp[i][j]为第j列中，满足从第k行至第i行为非递减的最小的k。dp[i][j] = A[i][j] >= A[i - 1][j] ? A[i - 1][j] : i;当且仅当k<=l时这一列满足条件。所以搜索每一列，只要找到这样满足条件的一列就输出Yes。k个查询，复杂度为O(mn + nk)。

上面的做法会超时。每次给出一组l和r的时候，矩阵中所有的列都会被取出，所以不用每次都遍历所有列，先在dp[i][j]中找出每一行最小的，只要最小的比l小那么这些列中一定有一列满足条件，复杂度为O(mn)。查询的时候只要O(k)。

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

int main() {
	int m, n; cin >> m >> n;
	vector<vector<int>> sheet(m, vector<int>(n));
	vector<vector<int>> dp(m, vector<int>(n));
	for (int i = 0; i < m; i++) {
		for (int j = 0; j < n; j++) {
			cin >> sheet[i][j];
		}
	}
	for (int i = 0; i < n; i++) {
		dp[0][i] = 0;
	}
	for (int i = 1; i < m; i++) {
		for (int j = 0; j < n; j++) {
			if (sheet[i][j] >= sheet[i - 1][j]) {
				dp[i][j] = dp[i - 1][j];
			}
			else {
				dp[i][j] = i;
			}
		}
	}
	vector<int> dp2(m, 0);
	for (int i = 0; i < m; i++) {
		dp2[i] = dp[i][0];
		for (int j = 1; j < n; j++) {
			if (dp[i][j] < dp2[i]) {
				dp2[i] = dp[i][j];
			}
		}
	}
	int k; cin >> k;
	for (int i = 0; i < k; i++) {
		int l, r; cin >> l >> r;
		bool f = dp2[r - 1] <= l - 1;
		if (f) printf("Yes\n");
		else printf("No\n");
	}
}