POJ1979
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).
题意解析：
这个题的意思是说给定一个W*H的矩形，里面的值为“.” "#"，其中“.”代表可到达的，“#”表示障碍，某人在一个“@”的起始点，求他所能到达的格子有多少个（包括第一所占的格子）。
思路：
1）将这个矩形根据值进行初始化，用map数组表示，“.”代表可到达的，用0表示，“#”表示障碍用1表示。记录起始点。
2）用一个数组visited存储位置的访问情况
3）进行一个简单的dfs，从起始点向四个方向分别进行dfs，如果可以到达（值为0），则将该位置的标志位visited标记为1.同时将结果数加1.
代码如下：
#include <stdio.h>
#define MAX_H 20
int map[MAX_H][MAX_H];
int visited[MAX_H][MAX_H];
int height;
int width;
int resultCount = 0;
int arr[4][2] = {{0,1},{1,0},{0,-1},{-1,0}}; //right down left up
int dfs(int x, int y);
int main()
{
//freopen("input.txt","r",stdin);
int i = 0;
int j = 0;
char tempChar;
scanf("%d %d",&width,&height);
int startWidth = 0;
int startHeight = 0;
while(!(width==0 && height==0))
{
resultCount = 1;
for(i=0;i<height;i++)
{
for(j=0;j<width;j++)
{
visited[i][j] = 0;
scanf(" %c",&tempChar);
if(tempChar==‘.‘)
{
map[i][j] = 0;
}
else if(tempChar==‘#‘)
{
map[i][j] = 1;
}
else if(tempChar==‘@‘)
{
map[i][j] = 2;
startWidth = j;
startHeight = i;
}
}
}
dfs(startHeight,startWidth);
printf("%d\n",resultCount);
scanf(" %d %d",&width,&height);
}
return 0;
}
int dfs(int x, int y)
{
int i = 0;
int tempx = 0;
int tempy = 0;
for(i=0;i<4;i++)
{
tempx = x + arr[i][0];
tempy = y + arr[i][1];
if(tempx>=0 && tempx<height && tempy>=0 && tempy<width)
{
if(map[tempx][tempy]==0 && visited[tempx][tempy]==0)
{
visited[tempx][tempy] = 1;
resultCount++;
dfs(tempx,tempy);
}
}
}
return 0;
}