HDU - 5706 GirlCat （dfs）

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本题是一道编程挑战题目，要求在给定的字符矩阵中找出并计数特定字符串图案girl和cat出现的次数。通过深度优先搜索算法遍历矩阵，寻找符合定义的图案。

GirlCat

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1020 Accepted Submission(s): 642

Problem Description

As a cute girl, Kotori likes playing ``Hide and Seek'' with cats particularly.
Under the influence of Kotori, many girls and cats are playing ``Hide and Seek'' together.
Koroti shots a photo. The size of this photo is

, each pixel of the photo is a character of the lowercase(from `a' to `z').
Kotori wants to know how many girls and how many cats are there in the photo.

We define a girl as -- we choose a point as the start, passing by 4 different connected points continuously, and the four characters are exactly ``girl'' in the order.
We define two girls are different if there is at least a point of the two girls are different.
We define a cat as -- we choose a point as the start, passing by 3 different connected points continuously, and the three characters are exactly ``cat'' in the order.
We define two cats are different if there is at least a point of the two cats are different.

Two points are regarded to be connected if and only if they share a common edge.

Input

The first line is an integer

which represents the case number.

As for each case, the first line are two integers

and

, which are the height and the width of the photo.
Then there are

lines followed, and there are

characters of each line, which are the the details of the photo.

It is guaranteed that:

is about 50.

Output

As for each case, you need to output a single line.
There should be 2 integers in the line with a blank between them representing the number of girls and cats respectively.

Please make sure that there is no extra blank.

Sample Input

  
  
   
   3
1 4
girl
2 3
oto
cat
3 4
girl
hrlt
hlca

Sample Output

Source

"巴卡斯杯" 中国大学生程序设计竞赛 - 女生专场

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题意：给一个 n * m 的矩阵，问其中有多少个 "girl" 和 "cat"

#include <bits/stdc++.h>
using namespace std;

int t, n, m;
char a[1003][1003];
int ans1, ans2;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};

int dfs(int i, int j, int k, int f){
    for(int t = 0; t < 4; t++){
        int ii = i + dx[t];
        int jj = j + dy[t];
        if(ii >= 0 &&ii < n && jj >= 0 && jj < m){
            if(f){
                if(k == 1 && a[ii][jj] == 'i'){
                    dfs(ii, jj, 2, f);
                }
                if(k == 2 && a[ii][jj] == 'r'){
                    dfs(ii, jj, 3, f);
                }
                if(k == 3&& a[ii][jj] == 'l'){
                    ans1++;
                }
            }
            else {
                if(k == 1 && a[ii][jj] == 'a'){
                    dfs(ii, jj, 2, f);
                }
                if(k == 2 && a[ii][jj] == 't'){
                    ans2++;
                }
            }
        }
    }
}

int main(){
    scanf("%d", &t);
    while(t--){
        scanf("%d%d", &n, &m);
        for(int i = 0; i < n; i++){
            scanf("%s", a[i]);
        }
        ans1 = 0, ans2 = 0;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                if(a[i][j] == 'g'){
                    dfs(i, j, 1, 1);
                }
                if(a[i][j] == 'c'){
                    dfs(i, j, 1, 0);
                }
            }
        }
        printf("%d %d\n", ans1, ans2);
    }
}