Counting Sheep
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 9 Accepted Submission(s) : 8
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Problem Description
A 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.
Input
The 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.
Output
For 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
0 < T <= 100
0 < H,W <= 100
Sample Input
2 4 4 #.#. .#.# #.## .#.# 3 5 ###.# ..#.. #.###
Sample Output
6 3
思路
连在一起的是一个羊群，让输出羊群的个数。
先找到一只羊，然后搜索所有和他属于一个羊群的羊，并把该羊群的所有点标志为已搜索。
第一次搜索结束之后，res++ ，接着，从另一个未被搜过的羊开始找另一个羊群。
......循环
直到找到这片区域。
深搜代码：
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxn = 100 + 10;
int dir[8] = {1,0,-1,0,0,1,0,-1};
char a[maxn][maxn];
bool visited[maxn][maxn];
int h,w,res;
void dfs(int n, int m){
visited[n][m] = true;
for (int i = 0; i < 8; i += 2) {
if ((n+dir[i]) >= 0 && (m+dir[i+1]) >= 0 && (n+dir[i]) < h && (m+dir[i+1]) < w
&& !visited[n+dir[i]][m+dir[i+1]] && a[n+dir[i]][m+dir[i+1]] == '#') {
dfs(n+dir[i], m+dir[i+1]);
}
}
}
int main() {
int t;
scanf("%d", &t);
while(t--) {
scanf("%d%d", &h, &w);
memset(a, 0, sizeof(a));
memset(visited, false, sizeof(visited));
res = 0;
for (int i = 0; i < h; i++) {
scanf("%s", a[i]);
}
for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) {
if (!visited[i][j] && a[i][j] == '#') {
res++;
dfs(i, j);
}
}
}
printf("%d\n", res);
}
return 0;
}广搜代码：
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
struct node{
int x;
int y;
node(int xx, int yy):x(xx),y(yy){}
};
const int maxn = 100 + 10;
int dir[8] = {1,0,-1,0,0,1,0,-1};
char a[maxn][maxn];
bool visited[maxn][maxn];
int h,w,res;
void bfs(int n, int m){
queue<node> q;
q.push(node(n,m));
res++;
while(!q.empty()){
node p = q.front();
q.pop();
int x = p.x;
int y = p.y;
for (int i = 0; i < 8; i += 2) {
if (x + dir[i] < h && y + dir[i+1] < w && (y + dir[i+1]) >= 0 && (x + dir[i]) >= 0
&& !visited[x + dir[i] ][ y + dir[i+1] ] && a[x + dir[i]][y + dir[i+1]] == '#') {
q.push(node(x + dir[i],y + dir[i+1]));
visited[x + dir[i]][y + dir[i+1]] = true;
}
}
}
}
int main() {
int t;
scanf("%d", &t);
while(t--) {
scanf("%d%d", &h, &w);
memset(a, 0, sizeof(a));
memset(visited, false, sizeof(visited));
res = 0;
for (int i = 0; i < h; i++) {
scanf("%s", a[i]);
}
for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) {
if (!visited[i][j] && a[i][j] == '#') {
bfs(i, j);
}
}
}
printf("%d\n", res);
}
return 0;
}