【DFS深搜初步】HDOJ-2952 Counting Sheep、NYOJ-27 水池数目

【题目链接：HDOJ-2952】

Counting Sheep

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2476    Accepted Submission(s): 1621

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】

　　深搜：就是把每种可能都枚举出来，直到找到符合条件的可能。

#include<iostream>
#include<cstring>
using namespace std;
const int MAXN = 105;
int Map[MAXN][MAXN] = {0};
int vis[MAXN][MAXN] = {0};
int dfs(int a,int b){
    if(Map[a][b] == 0 || vis[a][b] == 1) return 0;
    vis[a][b] = 1;
    //环顾四周 
    dfs(a - 1,b); //下
    dfs(a + 1,b); //上
    dfs(a,b - 1); //左
    dfs(a,b + 1); //右 
    return 0;
}
int main(){
    int n;
    cin >> n;
    while(n--){
        int a,b,i,j,sum = 0;
        memset(Map,0,sizeof(Map));
        memset(vis,0,sizeof(vis));
        cin >> a >> b;
        for(i = 0;i < a;i++){      
            for(j = 0;j < b;j++){
                char ac;
                cin >> ac; 
                if(ac == '#')
                    Map[i][j] = 1;
                else Map[i][j] = 0;                                    
            }            
        }                
        for(i = 0;i < a;i++)
        for(j = 0;j < b;j++){
            if(Map[i][j] == 0 || vis[i][j] == 1)
                continue;
            else{
                
                dfs(i,j);
                sum++;
            }
        }
        cout << sum << endl; 
    }
    return 0;
}

【题目链接：NYOJ-27】

　　可以说两题完全相似。

 
#include<iostream>
#include<cstring>
using namespace std;
const int MAXN = 105;
int Map[MAXN][MAXN] = {0};
int vis[MAXN][MAXN] = {0};
int dfs(int a,int b){
    if(Map[a][b] == 0 || vis[a][b] == 1) return 0;
    vis[a][b] = 1;
    //环顾四周 
    dfs(a - 1,b); //下
    dfs(a + 1,b); //上
    dfs(a,b - 1); //左
    dfs(a,b + 1); //右 
    return 0;
}
int main(){
    int n;
    cin >> n;
    while(n--){
        int a,b,i,j,sum = 0;
        memset(Map,0,sizeof(Map));
        memset(vis,0,sizeof(vis));
        cin >> a >> b;
        for(i = 1;i <= a;i++)
        for(j = 1;j <= b;j++){
                cin >> Map[i][j];
        }
        for(i = 1;i <= a;i++)
        for(j = 1;j <= b;j++){
            if(Map[i][j] == 0 || vis[i][j] == 1)
                continue;
            else{
                sum++;
                dfs(i,j);
            }
        }
        cout << sum << endl; 
    }
    return 0;
}