【CodeForces 777C】Alyona and Spreadsheet

Alyona and Spreadsheet

Problem Description

Duringthe lesson small girl Alyona works with one famous spreadsheet computer
programand learns how to edit tables.

Nowshe has a table filled with integers. The table consists ofnrows
andmcolumns.
Byai, jwe
will denote the integerlocated at thei-th
row and the j-th
column. We say that thetable is sorted in non-decreasing order in the columnjifai, j ≤ ai + 1, jfor
allifrom1ton - 1.

Teacher gave Alyonaktasks.
For each of thetasks two integerslandrare
given and Alyona has toanswer the following question: if one keeps the rows from l to r inclusive
and deletes allothers, will the table be sorted in non-decreasing order in at least onecolumn? Formally, does there exist suchjthatai, j ≤ ai + 1, jfor
allifromltor - 1inclusive.

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

Input

The first line of the input contains two positive integersnandm(1 ≤ n·m ≤ 100 000)
—the number of rows and the number of columns in the table respectively. Notethat 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 Iof
these lines stands forai, j(1 ≤ ai, j ≤ 109).

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

Thei-th
of the next k lines
contains two integers liandri(1 ≤ li ≤ ri ≤ n).

Output

Print "Yes"
tothe i-th
line of theoutput if the table consisting of rows from litoriinclusiveis
sorted in non-decreasing order in at least one column. Otherwise, print "No".

题目大意：

给出n*m的矩阵，k次查询，如果第l行到r行至少有一列是非递减的，则输出Yes，否则输出No

思路：

预处理每行能到达最大行是多少，再O(1)查询，注意把每列能到达的最大行保存一下，减少循环次数（否则会T）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define manx 100005
using namespace std;
int c[manx],d[manx];
int main()
{
int n,m;

scanf("%d%d",&n,&m);
int a[n+5][m+5]={0};
for (int i=1; i<=n; i++)
for (int j=1; j<=m; j++)
scanf("%d",&a[i][j]);
int k,cnt=1;
for (int i=1; i<=n; i++){
for (int j=1; j<=m; j++){
k=i;
if(d[j] >= i) k=d[j];   //优化
else
while(a[k+1][j] >= a[k][j] && k<n) k++;
d[j]=k;                //d the max of every column
c[cnt]=max(k,c[cnt]);  //c the max of every row
}
cnt++;
}
int q,x,y;
scanf("%d",&q);
for (int i=1; i<=q; i++){
scanf("%d%d",&x,&y);
if(c[x] >= y) printf("Yes\n");
else printf("No\n");
}
return 0;
}