HDU 2952 Counting Sheep (深度搜索)

Counting Sheep

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1753    Accepted Submission(s): 1143

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

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

 

Sample Output

63

 

Source

IDI Open 2009

 

Recommend

gaojie

题意：

有一块大草原,草原上有草有羊,“#”符号表示羊,“.”符号表示草。需要回答在这个大草坪中有几个羊群。
羊群的定义：某羊群中的任意两只羊都可通过相邻边联系在一起，当然也可能是仅有一只羊。

思路：

可使用深度搜索的方法解题。

/*************************************************************************> File Name: JJ.cpp> Author: BSlin> Mail:  > Created Time: 2013年09月26日 星期四 17时44分42秒 ************************************************************************/#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cstdlib>#include <cmath>#include <algorithm>#include <iterator>#include <vector>#include <map>#include <set>#include <stack>#include <queue>#define MP make_pair#define INF (1<<30)#define PI acos(-1.0)#define esp 1e-8const int dx[4]={0,0,0,0};using namespace std;#define read freopen("in.txt","r",stdin)#define write freopen("out.txt","w",stdout)#if defined (_WIN32) || defined (__WIN32) || defined (WIN32) || defined (__WIN32__)#define LL __int64#define LLS "%" "I" "6" "4" "d"#else#define LL long long#define LLS "%" "l" "l" "d"#endifchar map1[110][110];int n,m;int to[4][2] = {0,1,-1,0,0,-1,1,0};bool inmap(int x, int y) {    if(x >= 1 && x <= n && y >= 1 && y <= m) return true;    return false;}void dfs(int x, int y) {    int i,nowx,nowy;    for(i=0; i<4; i++) {        nowx = x + to[i][0];        nowy = y + to[i][1];        if(!inmap(nowx,nowy) || map1[nowx][nowy] == '.') continue;        map1[nowx][nowy] =  '.';        dfs(nowx,nowy);    }}int main(int argc, char** argv) {    //read;    int t,i,j,k,cnt;    scanf("%d",&t);    while(t--) {        scanf("%d %d\n",&n,&m);        cnt = 0;        k = 0;        for(i=1; i<=n; i++) {            for(j=1; j<=m; j++) {                scanf("%c",&map1[i][j]);            }            getchar();        }        for(i=1; i<=n; i++) {            for(j=1; j<=m;j++) {                if(map1[i][j] == '#') {                    cnt++;                    map1[i][j] = '.';                    dfs(i,j);                }            }        }        printf("%d\n",cnt);    }    return 0;}