HDU 5706 GirlCat DFS解决

GirlCat

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

Total Submission(s): 383    Accepted Submission(s): 261

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

 

Recommend

liuyiding

 

    题意：从输入的字符块中找出连续的"gril""cat"的个数。

    分析：DFS ，BFS都能过。实现保存girl数组和cat数组，DFS过程中迭代结果数组下标即可。比完看题解，甚至暴力if循环都能过……见AC代码：

#include<stdio.h>#include<string.h>using namespace std;const int maxn = 1005;char girl[4]= {'g','i','r','l'};char cat[3]= {'c','a','t'};char map[maxn][maxn];int vis[maxn][maxn];int n,m;int numgirl,numcat;void dfs(int x,int y,int z){if(x<0||x>=n||y<0||y>=m||map[x][y]!=girl[z])return;vis[x][y]=1;if(z==3){numgirl++;return;}dfs(x+1,y,z+1);dfs(x,y+1,z+1);dfs(x-1,y,z+1);dfs(x,y-1,z+1);}void dfs2(int x,int y,int z){if(x<0||x>=n||y<0||y>=m||map[x][y]!=cat[z])//y> m  不是nreturn;vis[x][y]=1;if(z==2){numcat++;return;}dfs2(x+1,y,z+1);dfs2(x,y+1,z+1);dfs2(x-1,y,z+1);dfs2(x,y-1,z+1);}int main(){int T;scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);numgirl=0,numcat=0;memset(vis,0,sizeof(vis));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(!vis[i][j]){if(map[i][j]=='g')dfs(i,j,0);else if(map[i][j]=='c')dfs2(i,j,0);}printf("%d %d\n",numgirl,numcat);}}

    DFS题还是需要多敲。

    特记下，以备后日回顾。