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
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
using namespace std;
struct node
{
int x;
int y;
int z;
} q[1000100];
int map[1001][1001];
int main()
{
int n,m,k;
while(scanf("%d%d%d",&n,&m,&k)!=EOF)
{
int x,y;
int flag = 0;
memset(map,0,sizeof(map));
for(int i=1; i<=k; i++)
{
/*for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
printf("%d ",map[i][j]);
}
printf("\n");
}*/
scanf("%d%d",&x,&y);
map[x][y] = 1;
if(flag == 0)
{
if(x+1<=n && y+1 <=m && map[x][y+1] == 1 && map[x+1][y+1] == 1 && map[x+1][y] == 1)
{
flag = i;
}
else if(x-1>0 && y-1>0 && map[x-1][y] == 1 && map[x][y-1] == 1 && map[x-1][y-1] == 1)
{
flag = i;
}
else if(x+1<=n && y-1>0 && map[x+1][y] == 1 && map[x][y-1] == 1 && map[x+1][y-1] == 1)
{
flag = i;
}
else if(x-1>0 && y+1<=m && map[x-1][y] == 1 && map[x][y+1] == 1 && map[x-1][y+1] == 1)
{
flag = i;
}
}
}
printf("%d\n",flag);
}
return 0;
}→Judgement Protocol
1, time: 62 ms., memory: 15668 KB, exit code: 0, checker exit code: 0, verdict: OK
Input
2 2 4
1 1
1 2
2 1
2 2
Output
4
Answer
4
Checker Log
ok answer is '4'
2, time: 31 ms., memory: 15664 KB, exit code: 0, checker exit code: 0, verdict: OK
Input
2 3 6
2 3
2 2
1 3
2 2
1 2
1 1
Output
5
Answer
5
Checker Log
ok answer is '5'
3, time: 0 ms., memory: 15664 KB, exit code: 0, checker exit code: 0, verdict: OK
Input
5 3 7
2 3
1 2
1 1
4 1
3 1
5 3
3 2
Output
0
Answer
0
Checker Log
ok answer is '0'
4, time: 46 ms., memory: 15668 KB, exit code: 0, checker exit code: 0, verdict: OK
Input
3 3 11
2 1
3 1
1 1
1 3
1 2
2 3
3 3
3 2
2 2
1 3
3 3
Output
9
Answer
9
Checker Log
ok answer is '9'
