HDU - 5706 GirlCat （dfs）

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 n\times m, 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 T which represents the case number.

As for each case, the first line are two integers n and m, which are the height and the width of the photo.
Then there are n lines followed, and there are m characters of each line, which are the the details of the photo.

It is guaranteed that:
T is about 50.
1\leq n\leq 1000.
1\leq m\leq 1000.
\sum (n\times m)\leq 2\times 10^6.

 

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

1 0 0 2 4 1

 

Source

"巴卡斯杯" 中國大學生程序設計競賽 - 女生專場

 

Recommend

liuyiding

題意：給一個 n * m 的矩陣，問其中有多少個 "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);
    }
}