Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 19731 Accepted Submission(s): 11991
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
Source
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1312
分析：看到ZERO已经写了DFS，弱鸡也得补上此题，来一发！感谢梅大大的支持,代码的模版来自梅大大,我只是对主要步骤做了注释而已！
下面给出AC代码：
#include <bits/stdc++.h>
using namespace std;
int _x[]={,,-,};
int _y[]={,-,,};//相当于从第一象限开始顺时针转一周，对应的坐标轴上的点分别为(1,0),(0,-1),(-1,0),(0,1),就是数学意义上的方向向量
char str[][];
bool flag[][];//去判断点是否被标记过,DFS搜索一遍,被标记过记改点为1!
int n,m,ans;
void DFS(int x,int y)
{
for(int ii=;ii<;ii++)
{
int i=x+_x[ii];
int j=y+_y[ii];
if(i<n&&j<m&&i>=&&j>=&&flag[i][j]==&&str[i][j]=='.')//边界条件
{
ans++;
flag[i][j]=;
DFS(i,j);
}
}
}
int main()
{
while(scanf("%d%d",&m,&n)!=EOF)
{
if(m==&&n==)
break;
int x,y;
memset(flag,,sizeof(flag));
for(int i=;i<n;i++)
scanf("%s",str[i]);
for(int i=;i<n;i++)
{
for(int j=;j<m;j++)
{
if(str[i][j]=='@')//找到起始点
{
x=i;
y=j;
}
}
}
ans=;
flag[x][y]=;
DFS(x,y);
printf("%d\n",ans+);//第一个位置也算一个，所以+1
}
return ;
}