Codeforces #401 (Div. 2) C. Alyona and Spreadsheet （ XJBX

C. Alyona and Spreadsheet

题目描述

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 ai, 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 ai, j ≤ ai + 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 ai, j ≤ ai + 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!

输入

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 ai, j (1 ≤ ai, 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 li and ri (1 ≤ li ≤ ri ≤ n).

Output

输出

Print “Yes” to the i-th line of the output if the table c

样例

inputCopy
5 4
1 2 3 5
3 1 3 2
4 5 2 3
5 5 3 2
4 4 3 4
6
1 1
2 5
4 5
3 5
1 3
1 5
output
Yes
No
Yes
Yes
Yes
No
Note
In the sample, the whole table is not sorted in any column. However, rows 1–3 are sorted in column 1, while rows 4–5 are sorted in column 3.

题意

给一个行列式问 然后 给两个数字行数问这个中间的列数有没有非递减的列数
本质是一个打表记忆的过程 类似记忆化？ 将每个行数可以向下的最大的列数纪录下去

AC代码

#include<bits/stdc++.h>

using namespace std;

#define LL long long
const int MAXN= 1e5+11;
int dp[MAXN];
int main()
{
    ios::sync_with_stdio(false);
    int n, m, q, x;
    cin >> n >> m;
    //vector<int> v[n+10];
    int v[n+10][m+10];
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            //cin >> x;
            //v[i].push_back(x);
            cin >> v[i][j];
        }

    }
    for(int i = 0; i < n; i++) dp[i] = i;
    for(int j = 0; j < m; j++) {
        int k = 0;
        for(int i = 1; i < n; i++) {
            if(v[i][j] >= v[i-1][j])
                dp[k] = max(dp[k], i);
            else
                k = i;
        }

    }
    for(int i = 1; i < n; i++)
        dp[i] = max(dp[i],dp[i-1]);
    cin >> q;
    while(q--) {
        int l, r;
        cin >> l >> r;
        l--, r--;
        if(dp[l] >= r)
            cout << "Yes\n";
        else
            cout << "No\n";
    }
    return 0;
}