hdu 5706 GirlCat【暴力DFS】

GirlCat

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

Total Submission(s): 50    Accepted Submission(s): 42

[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
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

[align=left]Sample Input[/align]
3

1 4

girl

2 3

oto

cat

3 4

girl

hrlt

hlca

[align=left]Sample Output[/align]
1 0

0 2

4 1

题目大意：有n*m个格子，从某一个格子出发，能向周围四个方向走，如果能够走到的地方能够组成单词girl，或者是cat都进行计数，最终输出其个数。

思路：枚举每个点，如果是g，深搜来找单词girl，如果是c，深搜来找单词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，我刚看到这个题的时候 ，感觉好蛋疼，数量级很大啊，1000*1000的图还要递归找单词，其实不然，不会超时滴。

AC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};
int ans;
int n,m;
void Dfs(int x,int y)
{
for(int i=0;i<4;i++)
{
int xx=x+fx[i];
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);
}
if(a[x][y]=='r'&&a[xx][yy]=='l')ans++;
}
}
}
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')
{
ans=0;
Dfs(i,j);
output+=ans;
}
if(a[i][j]=='c')
{
ans=0;
Dfs2(i,j);
output2+=ans;
}
}
}
printf("%d %d\n",output,output2);
}
}