H - Alyona and Spreadsheet
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H - Alyona and Spreadsheet
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 mcolumns. 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 ifai,?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 rinclusive 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 alli 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 lito ri inclusive is sorted in non-decreasing order in at least one column. Otherwise, print "No".
Example
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
Yes
No
Yes
Yes
Yes
No
题意:第一行输入两个整数n,m(1?≤?n·m?≤?100?000),第二行输入一个n×m的矩阵,然后再输入一个整数k,接下来k行,每行两个整数r,l,
询问从r到l行,如果存在一列数是非递减的,就输出Yes,否则就输出No。
题解:1.从上到下非递减,则上面的数比下面的数大就满足不了。
2.用二维数组用的不好就会超时,所以用一维数组a【】来存储每一行的数
3. 从后往前推,用一维数组b【】来存储每一列能到达的最上行,在用c【】来存储每一行能到达的最上行
4.最后只需要比较c【r】与l的大小即可。详情请看代码。
代码:
#include<iostream> using namespace std; int a[100005],b[100005],c[100005]; int main() { int n,m,x,i,j,r,l,k; cin>>n>>m; for(i=1;i<=m;i++)//没个数能到的最上一行是第一行。 b[i]=1; for(i=1;i<=n;i++){ c[i]=i; //每个数都能到达所在的行。 for(j=1;j<=m;j++){ cin>>x; if(x<a[j])//如果x上面的数比x大,则不是非递增的,所以x不能到达桑拿一行。 b[j]=i; a[j]=x; //存储每一行的数。 if(b[j]<c[i]) //找每一列能到达的最上行。 c[i]=b[j]; } } cin>>k; while(k--){ cin>>r>>l; if(c[l]<=r) cout<<"Yes"<<endl; else cout<<"No"<<endl; } return 0; }
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