GirlCat

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

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 Twhich 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)≤2×10^6.
 
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" 中国大学生程序设计竞赛 - 女生专场 
 
题解：一张n*m的英文字母照片，要求找到cat和gril，字母组成只需要连续，可以转弯。
 
思路：先找到首字母g和c的下标，用DFS找到gril和cat。因为能够走到的地方要是上一个格子的字母的下一个字母，所以这里极大的限制了Dfs走的层数，也极大的限制了能够走到的地方的个数，所以是不会超时的。
 
代码： 
#include<stdio.h>  
#include<string.h>  
using namespace std;  
char a[1005][1005];  
int fx[4]={0,0,-1,1};        
int fy[4]={1,-1,0,0};       //平移坐标
int ans;  
int n,m;  
void Dfs(int x,int y)  
{  
    for(int i=0;i<4;i++)  
    {  
        int xx=x+fx[i];           //上下左右四个点搜索
        int yy=y+fy[i];  
        if(xx>=0&&xx<n&&yy>=0&&yy<m)  
        {  
            if(a[x][y]=='g'&&a[xx][yy]=='i'||a[x][y]=='i'&&a[xx][yy]=='r')  
            {  
                Dfs(xx,yy);         //上一个是g，找到了i就继续搜索，上一个是i，找到了r就继续搜索
            }  
            if(a[x][y]=='r'&&a[xx][yy]=='l')ans++;        如果找到了r且下一个是l，答案+1
        }  
    }  
}  
void Dfs2(int x,int y)  
{  
    for(int i=0;i<4;i++)  
    {  
        int xx=x+fx[i];  
        int yy=y+fy[i];  
        if(xx>=0&&x<n&&yy>=0&&yy<m)  
        {  
            if(a[x][y]=='c'&&a[xx][yy]=='a')  
            {  
                Dfs2(xx,yy);  
            }  
            if(a[x][y]=='a'&&a[xx][yy]=='t')  
            {  
                ans++;  
            }  
        }  
    }  
}  
int main()  
{  
    int t;  
    scanf("%d",&t);  
    while(t--)  
    {  
        scanf("%d%d",&n,&m);  
        for(int i=0;i<n;i++)  
        {  
            scanf("%s",a[i]);  
        }  
        int output=0;  
        int output2=0;  
        for(int i=0;i<n;i++)  
        {  
            for(int j=0;j<m;j++)  
            {  
                if(a[i][j]=='g')           //标记g的坐标，开始深搜girl
                {  
                    ans=0;  
                    Dfs(i,j);  
                    output+=ans;       
                }  
                if(a[i][j]=='c')         //标记c，深搜cat
                {  
                    ans=0;  
                    Dfs2(i,j);  
                    output2+=ans;  
                }  
            }  
        }  
        printf("%d %d\n",output,output2);  
    }  
}

第二发