Sicily 1918. Counting square

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本文介绍了一种算法，用于计算给定矩阵中符合特定条件的魔法方阵数量。魔法方阵要求边界元素均为1，内部0和1的数量相差不超过1，且大小至少为2x2。文章提供了完整的C++实现代码。

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1918. Counting square

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

There is a matrix of size R rows by C columns. Each element in the matrix is either `0' or `1'. A square is called magic square if it meets the following three conditions.

The elements on the four borders are all `1'.
Inside the square (excluding the elements on the borders), the number of 1's and the number of 0's are different at most by 1.
The size of the square is at least 2 by 2. Now given the matrix, please tell me how many magic square are there in the matrix.

Input

The input begins with a line containing an integer T , the number of test cases. Each case begins with two integers R , C (1 ≤ R, C ≤ 300) , representing the size of the matrix. Then R lines follow. Each contains C integers, either `0' or `1'. The integers are separated by a single space.

Output

For each case, output the number of magic square in a single line.

Sample Input

Sample Output

3 
2 
1
// Problem#: 1918
// Submission#: 3577951
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include <stdio.h>

short G[305][305];
short row1[305][305], col1[305][305];
int sum[305][305];
int h, w;
int ans;

void init() {
    int last1;
    for (int i = 0; i < h; i++) {
        last1 = -1;
        for (int j = w - 1; j >= 0; j--) {
            if (G[i][j] == -1) last1 = -1;
            else if (last1 == -1) last1 = j;
            row1[i][j] = last1;
        }
    }
    for (int i = 0; i < w; i++) {
        last1 = -1;
        for (int j = h - 1; j >= 0; j--) {
            if (G[j][i] == -1) last1 = -1;
            else if (last1 == -1) last1 = j;
            col1[j][i] = last1;
        }
    }
    sum[0][0] = G[0][0];
    for (int i = 1; i < h; i++) sum[i][0] = sum[i - 1][0] + G[i][0];
    for (int i = 1; i < w; i++) sum[0][i] = sum[0][i - 1] + G[0][i];
    for (int i = 1; i < h; i++) 
        for (int j = 1; j < w; j++) 
            sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + G[i][j];
}

inline int abs(int a) {if (a < 0) return -a; return a;}

inline void cal(int si, int sj) {
    int li = si + 1, lj = sj + 1;
    while (li < h && lj < w && G[si][lj] == 1 && G[li][sj] == 1) {
        if (lj <= row1[li][sj] && li <= col1[si][lj])
            if (abs(sum[li - 1][lj - 1] - sum[si][lj - 1] - sum[li - 1][sj] + sum[si][sj]) <= 1) ans++;
        lj++;
        li++;
    }
}

int main() {
    int caseNum;
    scanf("%d", &caseNum);
    while (caseNum--) {
        scanf("%d%d\n", &h, &w);
        for (int i = 0; i < h; i++) {
            for (int j = 0; j < w; j++) {
                scanf("%d", &G[i][j]);
                if (G[i][j] == 0) G[i][j] = -1;
            }
        }
        ans = 0;
        init();
        for (int i = 0; i < h; i++) for (int j = 0; j < w; j++) if (G[i][j] == 1) cal(i, j);
        printf("%d\n", ans);
    }
    return 0;
}