HDU 1312-Red and Black (DFS)
Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25613 Accepted Submission(s): 15453
Problem Description
There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can’t move on red tiles, he can move only on black tiles.
Write a program to count the number of black tiles which he can reach by repeating the moves described above.
Write a program to count the number of black tiles which he can reach by repeating the moves described above.
Input
The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.
There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.
’.’ - a black tile
’#’ - a red tile
’@’ - a man on a black tile(appears exactly once in a data set)
There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.
’.’ - a black tile
’#’ - a red tile
’@’ - a man on a black tile(appears exactly once in a data set)
Output
For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).
Sample Input
6 9
….#.
…..#
……
……
……
……
……
#@…#
.#..#.
11 9
.#………
.#.#######.
.#.#…..#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#…….#.
.#########.
………..
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
…@…
###.###
..#.#..
..#.#..
0 0
Sample Output
45
59
6
13
代码：
#include <iostream>
#include <cstring>
using namespace std;
int dir[2][4]={ {
1,-1,0,0} , {
0,0,1,-1} };//四个方向
int mp[21][21];//用来标记已经走过的点
int w,h,t;
char c[21][21];
void DFS(int x,int y)//DFS核心代码
{
mp[x][y]=1;
for(int i=0;i<4;i++)
{
int dx=x+dir[0][i];
int dy=y+dir[1][i];
if(c[dx][dy]=='.' && !mp[dx][dy])
{
t++;
DFS(dx,dy);//满足条件则递归再次执行DFS函数，直到所有点都走完
}
}
}
int main()
{
int i,j;
while(cin>>w>>h && w,h)
{
t=1;
memset(c,0,sizeof(c));
memset(mp,0,sizeof(mp));//数组清除(初始化为0)
for(i=0;i<h;i++)
for(j=0;j<w;j++)
{
cin>>c[i][j];
}
for(i=0;i<h;i++)
{
for(j=0;j<w;j++)
{
if(c[i][j]=='@')
{
break;
}
}
if(c[i][j]=='@') break;
}
DFS(i,j);
cout<<t<<endl;
}
return 0;
}
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