Largest Submatrix-最大子矩阵-HDU-2870

Largest Submatrix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 970    Accepted Submission(s): 470


Problem Description

Now here is a matrix with letter 'a','b','c','w','x','y','z' and you can change 'w' to 'a' or 'b', change 'x' to 'b' or 'c', change 'y' to 'a' or 'c', and change 'z' to 'a', 'b' or 'c'. After you changed it, what's the largest submatrix with the same letters
you can make?

 

Input

The input contains multiple test cases. Each test case begins with m and n (1 ≤ m, n ≤ 1000) on line. Then come the elements of a matrix in row-major order on m lines each with n letters. The input ends once EOF is met.

 

Output

For each test case, output one line containing the number of elements of the largest submatrix of all same letters.

 

Sample Input

2 4
abcw
wxyz

 

Sample Output

3

 

此题类似求最大全1子矩阵(参考这里：http://blog.csdn.net/gaotong2055/article/details/8960118)，只不过这里要分别求abc的。

#include <stdio.h>

char matrix[1001][1001];
int h[1005], left[1005], right[1001];
char ms[3][1001][1001]; // 转换后a,b,c 的矩阵
char c3[4] = "abc";
char chage(char a, char b) { //转换char a 到 char b
if (a == 'w' && (b == 'a' || b == 'b'))
return b;
if (a == 'x' && (b == 'b' || b == 'c'))
return b;
if (a == 'y' && (b == 'c' || b == 'a'))
return b;
if (a == 'z')
return b;
return a;
}
int main() {
int m, n;
//freopen("in.txt", "r", stdin);
while (scanf("%d %d", &m, &n) != EOF) {
getchar(); //读取行尾
for (int i = 0; i < m; i++) {
gets(matrix[i]);
}
int max = 0, tmp, tmax;

//循环3次），求（a,b,c）最大子矩阵
for (int k = 0; k < 3; k++) {
for (int i = 0; i < m; i++) {
for (int j = 1; j <= n; j++)
ms[k][i][j] = chage(matrix[i][j-1], c3[k]); //转换矩阵
}

//求高
for (int j = 0; j <= n; j++) {
h[j] = left[j] = right[j] = 0;
}

//求最大子面积
for (int i = 0; i < m; i++) {
for (int j = 1; j <= n; j++) {
if (ms[k][i][j] == c3[k])
h[j] += 1;
else
h[j] = 0;
}
h[0] = h[n+1] = -1;
for (int j = 1; j <= n; j++) {
tmp = j;
while (h[j] <= h[tmp - 1] ) {
tmp = left[tmp - 1] ;
}
if (tmp < 0)
tmp = 0;
left[j] = tmp;
}

// for (int j = 1; j <= n; j++)
// printf("%d ", left[j]);
// printf("\n");

for (int j = n ; j > 0; j--) {
tmp = j;
while (h[j] <= h[tmp + 1] ) {
tmp = right[tmp + 1] ;
}

right[j] = tmp;
tmax = (tmp - left[j] + 1) * h[j];
if(max < tmax){
max = tmax;
}
}

// for (int j = 1; j <= n; j++)
// printf("%d ", right[j]);
// printf("\n");
}
}

printf("%d\n",max);

}

return 0;
}