HDU 1312:Red and Black
Time Limit:1000MS Memory Limit:30000KB 64bit IO Format:%I64d & %I64u
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.
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)
The end of the input is indicated by a line consisting of two zeros.
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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
题解：本题还是DFS搜索（上下左右），只是增加一个计数器，计算可以走的地方的个数。
规定地图中有可通行的位置，也有不可通行的位置，已知起点，求起点的与它相连成一片的部分，在这道题里输出相连的位置的数目。
从起点开始，遍历每一个到达的点的四个方向（不再是八个），到达一个位置就将这个位置的字符变成不可走的‘#‘，并且计数+1。其实就是计数将可走变成不可走的操作进行了多少次。
这样可以不用担心走过了还会重复。
#include <iostream>
using namespace std;
char a[25][25];
int n,m,total;
int dr[4] = {0,1,0,-1};//行变化
int dc[4] = {1,0,-1,0};//列变化
//上面的原来一直不会用，知道的话非常方便
bool judge(int x,int y)
{
if(x<1 || x>n || y<1 || y>m)
return 1;
if(a[x][y]==‘#‘)
return 1;
return 0;
}
void dfs(int r,int c)
{
total++;
a[r][c]=‘#‘; //走过一次，“。”变为“#”,避免重复
for(int k=0; k<4; k++)
{
int lr = r + dr[k];
int lc = c + dc[k];
if(judge(lr,lc))
continue;
dfs(lr,lc);
}
}
int main()
{
while(cin>>m>>n&&m&&n)
{
int i,j,x,y;
total=0;
for(i=1; i<=n; i++)
for(j=1; j<=m; j++)
{
cin>>a[i][j];
if(a[i][j]==‘@‘) //这里必须用变量x，y
x=i,y=j;
}
dfs(x,y);
cout<<total<<endl;
}
return 0;
}