Largest Submatrix-最大子矩陣-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 DescriptionNow 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? 

InputThe 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. 

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

Sample Input

2 4abcwwxyz

 

Sample Output

3

 

此題類似求最大全1子矩陣(參考這裡：http://blog.csdn.net/gaotong2055/article/details/8960118)，只不過這裡要分別求abc的。

#include <stdio.h>char matrix[1001][1001];int h[1005], left[1005], right[1001];char ms[3][1001][1001]; // 轉換後a,b,c 的矩陣char c3[4] = "abc";char chage(char a, char b) { //轉換char a 到 char bif (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(); //讀取行尾for (int i = 0; i < m; i++) {gets(matrix[i]);}int max = 0, tmp, tmax;//迴圈3次），求（a,b,c）最大子矩陣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]); //轉換矩陣}//求高for (int j = 0; j <= n; j++) {h[j] = left[j] = right[j] = 0;}//求最大子面積for (int i = 0; i < m; i++) {for (int j = 1; j <= n; j++) {if (ms[k][i][j] == c3[k])h[j] += 1;elseh[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;}