Description
Vasya has the square chessboard of size n × n and m rooks. Initially the chessboard is empty. Vasya will consequently put the rooks on the board one after another.
The cell of the field is under rook‘s attack, if there is at least one rook located in the same row or in the same column with this cell. If there is a rook located in the cell, this cell is also under attack.
You are given the positions of the board where Vasya will put rooks. For each rook you have to determine the number of cells which are not under attack after Vasya puts it on the board.
Input
The first line of the input contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ min(100 000, n2)) — the size of the board and the number of rooks.
Each of the next m lines contains integers xi and yi (1 ≤ xi, yi ≤ n) — the number of the row and the number of the column where Vasya will put the i-th rook. Vasya puts rooks on the board in the order they appear in the input. It is guaranteed that any cell will contain no more than one rook.
Output
Print m integer, the i-th of them should be equal to the number of cells that are not under attack after first i rooks are put.
Examples
input
3 31 13 12 2
output
4 2 0
input
5 21 55 1
output
16 9
input
100000 1300 400
output
9999800001 我们把x,y记录一下，根据情况去行还是去列
#include<bits/stdc++.h>
using namespace std;
long long n,m;
long long sum;
long long px[100005],py[100005];
long long ans[100005];
int main()
{
long long x,y;
int pos;
cin>>n>>m;
pos=n;
sum=n*n;
for(int i=1; i<=m; i++)
{
cin>>x>>y;
if(px[x]==0&&py[y]==0)
{
n--;
pos--;
// ans[i]=n*pos;
// cout<<"A"<<endl;
}
else if(px[x]&&py[y]==0)
{
pos--;
// ans[i]=n*pos;
}
else if(px[x]==0&&py[y])
{
n--;
// ans[i]=n*pos;
}
px[x]=1;
py[y]=1;
cout<<pos*n<<endl;
}
return 0;
}
Note
On the picture below show the state of the board after put each of the three rooks. The cells which painted with grey color is not under the attack.
