codeforce 777 C. Alyona and Spreadsheet （线段树）

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本文介绍了一个使用线段树解决表格排序查询的问题。通过预处理每一列的排序状态，并利用线段树进行区间更新和点查询，高效解答针对特定行范围内的列是否保持非递减排列的询问。

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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.

题目大意：给出一个表格，和k组询问，询问l,r行之间有没有一列单调不减。

题解：线段树。

对于每一列扫一遍，对于以第i行为起点合法的序列最多能维持到哪一行。

线段树的区间修改，点查询。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define N 100003
using namespace std;
int n,m,tr[N*4],delta[N*4];
void pushdown(int now)
{
	if (delta[now]) {
		delta[now<<1]=max(delta[now<<1],delta[now]);
		delta[now<<1|1]=max(delta[now<<1|1],delta[now]);
		tr[now<<1]=max(tr[now<<1],delta[now]);
		tr[now<<1|1]=max(tr[now<<1|1],delta[now]);
		delta[now]=0;
	}
}
void qjchange(int now,int l,int r,int ll,int rr,int v)
{
	if (ll<=l&&r<=rr) {
		tr[now]=max(tr[now],v);
		delta[now]=max(delta[now],v);
		return;
	} 
	int mid=(l+r)/2;
	pushdown(now);
	if (ll<=mid) qjchange(now<<1,l,mid,ll,rr,v);
	if (rr>mid) qjchange(now<<1|1,mid+1,r,ll,rr,v);
}
int find(int now,int l,int r,int x)
{
	if (l==r) return tr[now];
	int mid=(l+r)/2;
	pushdown(now);
	if (x<=mid) return find(now<<1,l,mid,x);
	else return find(now<<1|1,mid+1,r,x);
}
int main()
{
	freopen("a.in","r",stdin);
	freopen("my.out","w",stdout);
	scanf("%d%d",&n,&m);
	int a[n+2][m+2];
	for (int i=1;i<=n;i++)
	 for (int j=1;j<=m;j++) scanf("%d",&a[i][j]);
	for (int j=1;j<=m;j++){
	   int l,r; l=r=1;
	   for (int i=2;i<=n;i++) 
	    if (a[i][j]<a[i-1][j]){
	   	 qjchange(1,1,n,l,r,r);
	   	 //cout<<l<<" "<<r<<endl;
	   	 l=r=i;
	   }
	   else r++;
	   qjchange(1,1,n,l,r,r);
	   //cout<<l<<" "<<r<<endl;
	}
	int q; scanf("%d",&q);// cout<<q<<endl;
	for (int i=1;i<=q;i++) {
		int l,r; scanf("%d%d",&l,&r);// cout<<l<<" "<<r<<endl;
		int t=find(1,1,n,l);// cout<<t<<endl;
		if (t>=r) printf("Yes\n");
		else printf("No\n");
	}
}