【DP+预处理|最长连续子序列】HDU-2870 Largest Submatrix

Largest Submatrix

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

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 4abcwwxyz

 

Sample Output

3

 

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题意：可以将n×m的矩阵中的w、x、y、z按照规则替换成a、b、c找到最大的一个子矩阵，使其全部由a或b或c构成。

思路：这道题和HDU-1505是一样的，只不过有三种子矩阵而已：

HDU1505： http://blog.csdn.net/j_sure/article/details/41732341

代码如下：

/** * ID: j.sure.1 * PROG: * LANG: C++ */#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <ctime>#include <cmath>#include <stack>#include <queue>#include <vector>#include <map>#include <set>#include <string>#include <climits>#include <iostream>#define Mem(f, x) memset(f, x, sizeof(f))#define Pb push_back#define LL long longusing namespace std;const int INF = 0x3f3f3f3f;const double eps = 1e-8;/****************************************/const int N = 1111;int n, m, mat[N][N];int u[3][N][N];int l[3][N], r[3][N];//w -> a, b//x -> b, c//y -> a, c//z -> a, b, cint main(){#ifdef J_Sure    freopen("000.in", "r", stdin);    //freopen("999.out", "w", stdout);#endif    while(~scanf("%d%d", &n, &m)) {        char str[N];        for(int i = 1; i <= n; i++) {            str[0] = '*';            scanf("%s", &str[1]);            for(int j = 1; j <= m; j++) {                if(str[j] == 'a' || str[j] == 'y' || str[j] == 'z' || str[j] == 'w') {                    u[0][i][j] = u[0][i-1][j] + 1;                }                else u[0][i][j] = 0;                if(str[j] == 'b' || str[j] == 'w' || str[j] == 'x' || str[j] == 'z') {                    u[1][i][j] = u[1][i-1][j] + 1;                }                else u[1][i][j] = 0;                if(str[j] == 'c' || str[j] == 'x' || str[j] == 'y' || str[j] == 'z') {                    u[2][i][j] = u[2][i-1][j] + 1;                }                else u[2][i][j] = 0;            }        }        int best[3] = {}, ans = 0;        for(int k = 0; k < 3; k++) {            for(int i = 1; i <= n; i++) {                u[k][i][0] = u[k][i][m+1] = -1;                for(int j = 1; j <= m; j++) {                    l[k][j] = r[k][j] = j;                    while(u[k][i][j] <= u[k][i][l[k][j]-1]) {                        l[k][j] = l[k][l[k][j]-1];                    }                }                for(int j = m; j >= 1; j--) {                    while(u[k][i][j] <= u[k][i][r[k][j]+1]) {                        r[k][j] = r[k][r[k][j]+1];                    }                }                for(int j = 1; j <= m; j++) {                    int cur = (r[k][j] - l[k][j] + 1)*u[k][i][j];                    best[k] = max(best[k], cur);                }            }            ans = max(ans, best[k]);        }        printf("%d\n", ans);    }    return 0;}