hdu 5706 GirlCat(深搜)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5706

——报考杭州电子科技大学！

GirlCat

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 132    Accepted Submission(s): 104

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×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≤n≤1000.
1≤m≤1000.
∑(n×m)≤2×106.

 

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

 

题目大意：根据输入的字符矩阵，分别找到girl和cat字符串的数量。

解题思路：找到一个起始点，直接进行搜索，查找接下去的字母。

详见代码。

#include <iostream>#include <cstdio>using namespace std;char Map[1010][1010];char ans[2][10]= {{"cat"},{"girl"}};int dir[4][2]= {0,1,0,-1,1,0,-1,0}; //4行2列int n,m,s1,s2;void dfs(int x,int y,int flag,int num)//flag用来标记我当前找的字符串是cat还是girl，num用来表示当前找到了是第几个字符。{    if (flag==1)//girl    {        if (num==3)//找到了girl的长度就加加        {            s1++;            return;        }        for (int i=0; i<4; i++)        {            int X=x+dir[i][0];            int Y=y+dir[i][1];            if (X>=0&&X<n&&Y>=0&&Y<m&&(Map[X][Y]==ans[flag][num+1]))//一个字符一个字符比较，看是否为我接下去要找的那个字符            {                dfs(X,Y,1,num+1);            }        }    }    if (flag==0)    {        if (num==2)        {            s2++;            return;        }        for (int i=0; i<4; i++)        {            int X=x+dir[i][0];            int Y=y+dir[i][1];            if (X>=0&&X<n&&Y>=0&&Y<m&&(Map[X][Y]==ans[flag][num+1]))            {                dfs(X,Y,0,num+1);            }        }    }}int main(){    int t;    scanf("%d",&t);    while (t--)    {        s1=s2=0;        scanf("%d%d",&n,&m);        for (int i=0; i<n; i++)        {            scanf("%s",Map[i]);        }        for (int i=0; i<n; i++)        {            for (int j=0; j<m; j++)            {                if (Map[i][j]=='g')                    dfs(i,j,1,0);                if (Map[i][j]=='c')                    dfs(i,j,0,0);            }        }        printf ("%d %d\n",s1,s2);    }    return 0;}