Hdu2952 counting sheep

Counting sheep

Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 1504 accepted submission (s): 981

Problem descriptiona while ago I had trouble sleeping. I used to lie awake, staring at the ceiling, for hours and hours. then one day my grandmother suggested I tried counting sheep after I 'd gone to bed. as always when my grandmother suggests things,
I decided to try it out. The only problem was, there were no sheep around to be counted when I went to bed.

Creative as I am, that wasn' t going to stop me. I sat down and wrote a computer program that made a grid of characters, where # represents a sheep, while. is grass (or whatever you like, just not sheep ). to make the counting a little more interesting, I also
Decided I Wanted To count flocks of sheep instead of single sheep. two sheep are in the same flock if they share a common side (Up, down, right or left ). also, if sheep a is in the same flock as sheep B, and sheep B is in the same flock as sheep C, then sheeps
A and C are in the same flock.

Now, I 've got a new problem. though counting these sheep actually helps me fall asleep, I find that it is extremely boring. to solve this, I 've decided I need another computer program that does the counting for me. then I'll be able to just start both these
Programs before I go to bed, and I'll sleep tight until the morning without any disturbances. I need you to write this program for me.

Inputthe first line of input contains a single number t, the number of test cases to follow.

Each test case begins with a line containing two numbers, H and W, the height and width of the sheep grid. then follows H lines, each containing W characters (either # Or .), describing that part of the grid.

 

Outputfor each test case, output a line containing a single number, the amount of sheep flock son that grid according to the rules stated in the Problem description.

Notes and constraints
Zero <t <= 100
0

Sample Input

24 4#.#..#.##.##.#.#3 5###.#..#..#.###

 

Sample output

63

 

Sourceidi open2009
 

Recommendgaojie

Solution: This question is a simple question of breadth-first search. It can be viewed as the graph's connectivity search. You can find several connected small graphs in the big graph.
Count sheep --> count the sheep and traverse the entire grassland. Once the sheep are found, the whole group is marked to avoid heavy data. Then, continue to traverse until the entire grassland is completely traversed.

# Include <cstdio> # include <cstring> # include <queue> using namespace STD; int n, m; char map [100] [100]; int sum; int dir [4] [2] = {, 0,-1,-}; struct node {int X; int y ;}; int inmap (int x, int y) {If (x> = 0 & x <n & Y> = 0 & Y <m) return true; return false;} void BFS (int x, int y) {int I; node u, v; queue <node> q; U. X = x; U. y = y; q. push (U); While (! Q. empty () {u = Q. front (); q. pop (); for (I = 0; I <4; I ++) {v. X = u. X + dir [I] [0]; V. y = u. Y + dir [I] [1]; If (inmap (v. x, V. y) & map [v. x] [v. y] = '#') // here there is a sheep that has never been counted {map [v. x] [v. y] = '*'; // This goat has already been counted. Next time, do not count it. Q. push (v) ;}}} int main () {int I, j; int t; scanf ("% d", & T); While (t --) {scanf ("% d", & N, & M); sum = 0; for (I = 0; I <n; I ++) scanf ("% s", map [I]); for (I = 0; I <n; I ++) {for (j = 0; j <m; j ++) {If (Map [I] [J] = '#') // discover sheep {sum ++; BFS (I, j ); // locate the whole group of sheep and mark them to avoid duplicates} printf ("% d \ n", sum);} return 0 ;}