Codeforces Round #401 (Div. 2) C Alyona and Spreadsheet —— 打表

题目链接：http://codeforces.com/contest/777/problem/C

 

C. Alyona and Spreadsheet

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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!

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 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

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

Example

input

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.

 

 

题解：

 

给出一个表格，给出l和r，问在l行和r行内（包括l行和r行），是否最少有一列，这列数随行数的递增，为非递减数列。

其中有k个l和r，k<=1000000, 情况这么大，每读入l r就对表格就行访问肯定会超时的，所以只能打表，读l r时直接访问。

其实怎么打表我是不会做的，所以还是看了通过的代码，发现他们都好聪明。

方法如下：用tt[j]现时记录在i行内，在aij以上（包括aij）有多少个非递减数列项，用maxn[i]记录在i行里tt[j]的最大值，即记录在i行内，哪一列的非递减数列项的个数最多。

于是在输入l r时，r-l+1即为需要判断的数列项个数， maxn[r]即为实际的数列项个数，如果maxn[r]>=r-l+1, 即表明l r行内存在非递减数列。

代码如下：

 

#include<stdio.h>  
#include<cstring>  
#define MAX(a,b) (a>b?a:b)  
int main()  
{  
    int n,m,k,l,r;  
    scanf("%d%d",&n,&m);  
    int a[n+5][m+5],maxn[n+5],tt[m+5];  
    memset(maxn,0,sizeof(maxn));  
    memset(tt,0,sizeof(tt));  
    for(int i = 0; i<n; i++)  
    for(int j = 0; j<m; j++)  
    {  
        scanf("%d",&a[i][j]);  
        if(i==0 || a[i][j]>=a[i-1][j]) tt[j]++;  
        else tt[j] = 1;  
        maxn[i] = MAX(maxn[i],tt[j]);  
    }  
    scanf("%d",&k);  
    for(int i = 0; i<k; i++)  
    {  
        scanf("%d%d",&l,&r);  
        if(r-l+1<=maxn[r-1]) puts("Yes");  
        else puts("No");  
    }  
    return 0;  
}