A. Pasha and Pixels
time limit per test
memory limit per test
input
output
Pasha loves his phone and also putting his hair up... But the hair is now irrelevant.
n row with mpixels in each row. Initially, all the pixels are colored white. In one move, Pasha can choose any pixel and color it black. In particular, he can choose the pixel that is already black, then after the boy's move the pixel does not change, that is, it remains black. Pasha loses the game when a 2 × 2
k moves, according to which he will paint pixels. Each turn in his plan is represented as a pair of numbers iand j, denoting respectively the row and the column of the pixel to be colored on the current move.
2 × 2
Input
n, m, k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 105) — the number of rows, the number of columns and the number of moves that Pasha is going to perform.
k lines contain Pasha's moves in the order he makes them. Each line contains two integers i and j (1 ≤ i ≤ n, 1 ≤ j ≤ m), representing the row number and column number of the pixel that was painted during a move.
Output
2 × 2
2 × 2 square consisting of black pixels is formed during the given k moves, print 0.
Sample test(s)
input
2 2 4 1 1 1 2 2 1 2 2
output
4
input
2 3 6 2 3 2 2 1 3 2 2 1 2 1 1
output
5
input
5 3 7 2 3 1 2 1 1 4 1 3 1 5 3 3 2
output
0
暴力即可
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (1000+10)
#define MAXK (100000+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n,m,k;
bool b[MAXN][MAXN]={0};
int main()
{
// freopen("Pixels.in","r",stdin);
// freopen(".out","w",stdout);
cin>>n>>m>>k;
For(i,k)
{
int x,y;
scanf("%d%d",&x,&y);
b[x][y]=1;
bool flag=0;
if (b[x-1][y]&&b[x-1][y-1]&&b[x][y-1]) flag=1;
if (b[x+1][y]&&b[x+1][y+1]&&b[x][y+1]) flag=1;
if (b[x-1][y]&&b[x-1][y+1]&&b[x][y+1]) flag=1;
if (b[x+1][y]&&b[x+1][y-1]&&b[x][y-1]) flag=1;
if (flag)
{
cout<<i<<endl;
return 0;
}
}
cout<<"0"<<endl;
return 0;
}
