HDU

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

31 4girl2 3otocat3 4girlhrlthlca

 

Sample Output

1 00 24 1

 

Source

"巴卡斯杯" 中国大学生程序设计竞赛 - 女生专场

 

Recommend

liuyiding

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);    }}