NYoj-27-pool quantity-DFS-BFS

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.
For each set of data, the number of rows m (0 <m <100) and number of columns n (0 <n <100) of the map are input first. Then, enter n numbers per line in the next m line, indicating whether there is water or no water here (1 indicates the pool, 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 05 51 1 1 1 00 0 1 0 0 0 0 01 1 1 0 00 0 1 1 1

Sample output

23

DFS:

# Include <cstdio> # include <cstring> # define N 200 # define M 200int map [N] [M] = {0}; void DFS (int I, int j) {if (map [I] [j-1]) {map [I] [j-1] = 0; DFS (I, j-1 );} if (map [I] [j + 1]) {map [I] [j + 1] = 0; DFS (I, j + 1 );} if (map [i-1] [j]) {map [i-1] [j] = 0; DFS (i-1, j );} if (map [I + 1] [j]) {map [I + 1] [j] = 0; DFS (I + 1, j) ;}} int main () {int t, n, m; int I, j, count; scanf ("% d", & t); while (t --) {scanf ("% d", & n, & m); count = 0; for (I = 1; I <= n; I ++) for (j = 1; j <= m; j ++) scanf ("% d", & map [I] [j]); for (I = 1; I <= n; I ++) for (j = 1; j <= m; j ++) if (map [I] [j]) {count ++; map [I] [j] = 0; DFS (I, j);} printf ("% d \ n", count);} return 0 ;}
Let's look at this again:
# Include <cstdio> # include <cstring> # define N 200 # define M 200int map [N] [M] = {0 }; int f [4] [2] = {-}, {0,-1}; void DFS (int I, int j) {int k; map [I] [j] = 0; for (k = 0; k <4; k ++) {if (map [I + f [k] [0] [j + f [k] [1] = 1) DFS (I + f [k] [0], j + f [k] [1]) ;}} int main () {int t, n, m; int I, j, count; scanf ("% d", & t); while (t --) {scanf ("% d", & n, & m ); count = 0; for (I = 1; I <= n; I ++) for (j = 1; j <= m; j ++) scanf ("% d", & map [I] [j]); for (I = 1; I <= n; I ++) for (j = 1; j <= m; j ++) if (map [I] [j]) {count ++; map [I] [j] = 0; DFS (I, j);} printf ("% d \ n", count);} return 0 ;}
BFS:
# Include <cstdio> # include <cstring> # include <algorithm> # include <queue> # define N 200 # define M 200 using namespace std; int map [N] [M] = {0}; int f [4] [2] = {-}, {0, -1 }}; struct pool {int a, B ;}; queue <pool> v; void BFS (int I, int j) {int k; pool p, x, y; p. a = I; p. B = j; v. push (p); map [I] [j] = 0; while (! V. empty () {x = v. front (); for (k = 0; k <4; k ++) {if (map [x. a + f [k] [0] [x. B + f [k] [1]) {map [x. a + f [k] [0] [x. B + f [k] [1] = 0; y. a = x. a + f [k] [0]; y. B = x. B + f [k] [1]; v. push (y) ;}} v. pop () ;}} int main () {int t, n, m; int I, j, count; scanf ("% d", & t ); while (t --) {scanf ("% d", & n, & m); count = 0; for (I = 1; I <= n; I ++) for (j = 1; j <= m; j ++) scanf ("% d", & map [I] [j]); for (I = 1; I <= n; I ++) for (j = 1; j <= m; j ++) if (map [I] [j]) {count ++; map [I] [j] = 0; BFS (I, j);} printf ("% d \ n", count) ;}return 0 ;}

NYoj-27-pool quantity-DFS-BFS