H - Alyona and Spreadsheet
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
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
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
Input5 4Output
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
題意:第一行輸入兩個整數n,m(1?≤?n·m?≤?100?000),第二行輸入一個n×m的矩陣,然後再輸入一個整數k,接下來k行,每行兩個整數r,l,
詢問從r到l行,如果存在一列數是非遞減的,就輸出Yes,否則就輸出No。
題解:1.從上到下非遞減,則上面的數比下面的數大就滿足不了。
2.用二維數組用的不好就會超時,所以用一維數組a【】來存儲每一行的數
3. 從後往前推,用一維數組b【】來存儲每一列能到達的最上行,在用c【】來存儲每一行能到達的最上行
4.最後只需要比較c【r】與l的大小即可。詳情請看代碼。
代碼:
#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++)//沒個數能到的最上一行是第一行。 b[i]=1; for(i=1;i<=n;i++){ c[i]=i; //每個數都能到達所在的行。 for(j=1;j<=m;j++){ cin>>x; if(x<a[j])//如果x上面的數比x大,則不是非遞增的,所以x不能到達桑拿一行。 b[j]=i; a[j]=x; //存儲每一行的數。 if(b[j]<c[i]) //找每一列能到達的最上行。 c[i]=b[j]; } } cin>>k; while(k--){ cin>>r>>l; if(c[l]<=r) cout<<"Yes"<<endl; else cout<<"No"<<endl; } return 0; }
H - Alyona and Spreadsheet