Number of nyoj 27 pools

 

Pool quantity time limit: 3000 MS | memory limit: 65535 kb difficulty: 4

Description

There are some small rivers and some lakes on the campus of Nanyang Institute of Technology. Now we take them all as pools. Suppose there is a map somewhere in our school, this map only identifies whether it is a pool. Now, your task is coming. Please use a computer to figure out several pools in the map.

 

Input

Enter an integer N in the first line, indicating that there are N groups of test data. Each group of data first enters the number of rows of the map (0 <m <100) and the number of columns N (0 <n <100). Then, enter n numbers for each row in the next M line, indicating whether there is water or no water (1 indicates that the pool is used here, 0 indicates the ground)

Output

Output The number of pools in the map. Note that, if the pool is still located next to each pool (up or down the four locations), they can be seen as the same pool.

Sample Input

23 41 0 0 0 0 0 1 11 1 1 05 51 1 1 1 00 0 1 0 10 0 0 0 01 1 1 0 00 0 1 1 1

Sample output

23

# Include <stdio. h> # include <string. h> int sum; int M, N; int e [100] [100]; // the two-dimensional array int next [4] [2] =,, 0,-1,-}; // defines a direction array void DFS (int I, Int J) // Deep Search {int X, Y; E [I] [J] = 0; // assign all connected pools 0; For (int K = 0; k <4; k ++) {x = I + next [k] [0]; y = J + next [k] [1]; if (x <0 | x> M-1 | Y <0 | Y> N-1) // you can determine whether to cross-border continue; else if (E [x] [Y] = 1) {DFS (x, y) ;}} return ;}int main () {int num; scanf ("% d", & num); While (Num --) {sum = 0; scanf ("% d", & M, & N ); for (INT I = 0; I <m; I ++) {for (Int J = 0; j <n; j ++) {scanf ("% d ", & E [I] [J]) ;}}for (INT I = 0; I <m; I ++) {for (Int J = 0; j <N; j ++) {If (E [I] [J] = 1) {sum ++; DFS (I, j );}}} printf ("% d \ n", sum);} return 0 ;}