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
題意:
給你一個n*m的矩陣,問 l 到 r 行是否有某一列是非遞減的(假設第 j 列滿足,則任意i>.=l&&i<=r s[i][j]>=s[i-1 ][j]總成立) 問k次;
思路:
開一個dp數組 dp[i[j]保存mp[i][j]向上幾位滿足條件,然後開一個一維數組保存第i行可以向上幾行滿足條件(一定要預處理,不然會TLE!!!) 然後O(1)次詢問即可.預處理真香!!!
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,m,x,s[100000];
vector<int> mp[100000],dp[100000];
scanf("%d %d",&n,&m);
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
scanf("%d",&x);
mp[i].push_back(x);
}
for(int j=0;j<m;j++)//第一行預處理爲一
dp[0].push_back(1);
for(int j=0;j<m;j++)
{
for(int i=1;i<n;i++)
{
if(mp[i][j]>=mp[i-1][j])
dp[i].push_back(dp[i-1][j]+1);
else
dp[i].push_back(1);
}
}
memset(s,0,sizeof s);
for(int i=0;i<n;i++)//預處理
for(int j=0;j<m;j++)
s[i]=max(s[i],dp[i][j]);//保存最大的那個
int k;
scanf("%d",&k);
while(k--)
{
int l,r;
bool flag=0;
scanf("%d%d",&l,&r);
if(s[r-1]>=(r-l+1))
printf("Yes\n");
else
printf("No\n");
}
return 0;
}