Maximum P1387 square, P1387 Square

Maximum P1387 square, P1387 Square
Description

Find the maximum square that does not contain 0 in a matrix containing only 0 and 1 of n * m, and the length of the output side.

Input/Output Format Input Format:

The first row of the input file is two integers, n, m (1 <= n, m <= 100), followed by n rows. Each row contains m numbers, which are separated by spaces, 0 or 1.

Output Format:

An integer with the side length of the largest square

Input and Output sample Input example #1:

4 40 1 1 11 1 1 00 1 1 01 1 0 1

Output sample #1:

2

1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <cmath> 5 using namespace std; 6 int map [101] [101]; 7 int dp [101] [101]; 8 int main () 9 {10 int n, m; 11 scanf ("% d", & n, & m ); 12 for (int I = 1; I <= n; I ++) 13 for (int j = 1; j <= m; j ++) 14 {15 scanf ("% d", & map [I] [j]); 16 if (map [I] [j]) 17 dp [I] [j] = 1; 18} 19 for (int I = 1; I <= n; I ++) 20 for (int j = 1; j <= m; j ++) 21 {22 if (map [I] [j]) 23 {24 dp [I] [j] = min (dp [I-1] [j], dp [I] [J-1]), dp [I-1] [J-1]) + 1; 25} 26} 27 int ans =-1; 28 for (int I = 1; I <= n; I ++) 29 {30 for (int j = 1; j <= m; j ++) 31 {32 ans = max (ans, dp [I] [j]); 33} 34} 35 printf ("% d", ans); 36 return 0; 37}AC 1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <cmath> 5 using namespace std; 6 int map [101] [101]; 7 int dp [80] [80] [80] [80]; 8 int main () 9 {10 int n, m; 11 scanf ("% d ", & n, & m); 12 for (int I = 1; I <= n; I ++) 13 for (int j = 1; j <= m; j ++) 14 {15 scanf ("% d", & map [I] [j]); 16 if (map [I] [j]) 17 dp [I] [j] [I] [j] = 1; 18} 19 20 for (int I = 1; I <= n; I ++) 21 {22 for (int j = 1; j <= m; j ++) 23 {24 if (map [I] [j]) 25 {26 for (int k = I; k <= n; k ++) 27 {28 for (int l = j; l <= m; l ++) 29 {30/* if (k-1 = 0) 31 dp [I] [j] [k] [l] = dp [I] [j] [k] [l] | (dp [I] [j] [k] [L-1] & map [k] [l]); 32 else if (L-1 = 0) 33 dp [I] [j] [k] [l] = dp [I] [j] [k] [l] |] [l] & map [k] [l]); 34 else */35 dp [I] [j] [k] [l] = dp [I] [j] [k] [l] | (dp [I] [j] [k-1] [l] | dp [I] [j] [k] [L-1]) & map [k] [l]); 36} 37} 38} 39 40} 41} 42 for (int I = 1; I <= n; I ++) 43 {44 for (int j = 1; j <= m; j ++) 45 {46 if (map [I] [j]) 47 {48 for (int k = I; k <= n; k ++) 49 {50 for (int l = j; l <= m; l ++) 51 {52/* if (k-1 = 0) 53 dp [I] [j] [k] [l] = dp [I] [j] [k] [l] | (dp [I] [j] [k] [L-1] & map [k] [l]); 54 else if (L-1 = 0) 55 dp [I] [j] [k] [l] = dp [I] [j] [k] [l] |] [l] & map [k] [l]); 56 else */57 if (dp [I] [j] [k-1] [l-1] & dp [I] [j] [k-1] [l] & dp [I] [j] [k] L-1] & map [k] [l] = 1) 58 dp [I] [j] [k] [l] = max (dp [I] [j] [k] [l], dp [I] [j] [k-1] [L-1] + 1); 59} 60} 61} 62 63} 64} 65 int ans =-1; 66 for (int I = 1; I <= n; I ++) 67 {68 for (int j = 1; j <= m; j ++) 69 {70 for (int k = 1; k <= n; k ++) 71 {72 for (int l = 1; l <= m; l ++) 73 {74 ans = max (ans, dp [I] [j] [k] [l]); 75} 76} 77} 78} 79 printf ("% d ", ans); 80 return 0; 81}Four-Dimensional 30