Round C APAC Test 2017 Problem B. Monster Path

Problem B. Safe Squares

Codejamon trainers are actively looking for monsters, but if you are not a trainer, these monsters could be really dangerous for you. You might want to find safe places that do not have any monsters!

Consider our world as a grid, and some of the cells have been occupied by monsters. We define a safe square as
a grid-aligned D × D square
of grid cells (with D ≥ 1) that does not contain any monsters. Your task is to find out how many safe squares (of any size)
we have in the entire world.

Input

The first line of the input gives the number of test cases, T. T test
cases follow. Each test case starts with a line with three integers, R, C,
and K. The grid has R rows
and Ccolumns, and contains K monsters. K more
lines follow; each contains two integers Ri andCi,
indicating the row and column that the i-th monster is in. (Rows are numbered from top to bottom, starting from 0; columns are numbered from left to right, starting from 0.)

Output

For each test case, output one line containing 

Case #x: y

,
where 

x

 is the test case number (starting from 1) and 

y

 is
the the total number of safe zones for this test case.

Limits

1 ≤ T ≤ 20.
(Ri, Ci) ≠ (Rj, Cj) for i ≠ j. (No two monsters are in the same grid cell.)
0 ≤ Ri < R, i from 1 to K
0 ≤ Ci < C, i from 1 to K

Small dataset

1 ≤ R ≤ 10.
1 ≤ C ≤ 10.
0 ≤ K ≤ 10.

Large dataset

1 ≤ R ≤ 3000.
1 ≤ C ≤ 3000.
0 ≤ K ≤ 3000.

Sample

Input	Output
2 3 3 1 2 1 4 11 12 0 1 0 3 0 4 0 10 1 0 1 9 2 0 2 4 2 9 2 10 3 4 3 10	Case #1: 10 Case #2: 51

The grid of sample case #1 is:

0 0 0

0 0 0

0 1 0

Here, 0 represents a cell with no monster, and 1 represents a cell with a monster. It has 10 safe squares: 8 1x1 and 2 2x2.
The grid of sample case #2 is:

0 1 0 1 1 0 0 0 0 0 1

1 0 0 0 0 0 0 0 0 1 0

1 0 0 0 1 0 0 0 0 1 1

0 0 0 0 1 0 0 0 0 0 1

Note that sample case #2 will only appear in the Large dataset. It has 51 safe squares: 32 1x1, 13 2x2, 5 3x3, and 1 4x4.
S[i][j] represents size of the square sub-matrix with all 1s including M[i][j] where M[i][j] is the rightmost and bottommost entry in sub-matrix.

Solution

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

typedef long long ll;

const int MAXN = 3000;
bool grid[MAXN][MAXN];
ll dp[MAXN][MAXN];

ll get_res(int R, int C)
{
ll res = 0;

for (int i = 0; i < R; i++)
{
dp[i][0] = !grid[i][0];
res += dp[i][0];
}

for (int j = 1; j < C; j++)
{
dp[0][j] = !grid[0][j];
res += dp[0][j];
}

for (int i = 1; i < R; i++)
for (int j = 1; j < C; j++)
{
if (!grid[i][j])
{
dp[i][j] = min(dp[i-1][j], min(dp[i][j-1], dp[i-1][j-1])) + 1;
res += dp[i][j];
}
else
dp[i][j] = 0;
}

return res;
}

int main(int argc, char* argv[])
{
#ifndef TEST
if (argc != 2)
{
cout << "Invalid input" << endl;
return 1;
}

string input = argv[1];
string output = input.substr(0, input.length() - 2) + "out";
freopen(input.c_str(), "r", stdin);
freopen(output.c_str(), "w", stdout);
#endif

int T, R, C, K, Ri, Ci;
cin >> T;
for (int i = 1; i <= T; i++)
{
memset(grid, 0, MAXN * MAXN * sizeof(bool));
memset(dp, 0, MAXN * MAXN * sizeof(ll));

cin >> R >> C >> K;
for (int j = 0; j < K; j++)
{
cin >> Ri >> Ci;
grid[Ri][Ci] = 1;
}

ll res = get_res(R, C);
cout << "Case #" << i << ": " << res << endl;
}

fclose(stdin);
fclose(stdout);

return 0;
}

Note

dp[i][j] represents width of the maximum square sub-matrix with all 0s including grid[i][j] where grid[i][j] is the rightmost and bottommost entry in sub-matrix.

Reference

https://code.google.com/codejam/contest/6274486/dashboard#s=p1
http://codeforces.com/blog/entry/47185