标签：poj1979 poj 1979 red and black
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Red and Black
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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0 Sample Output 45 59 6 13 Source |
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直接附上代码吧。很简单的深搜
#include <stdio.h>
#include <string.h>
char map[25][25];
int dir[4][2]={1,0,-1,0,0,1,0,-1},sum,n,m;
bool limit(int x,int y)//是否出界
{
if(x>=0&&x<n&&y>=0&&y<m)
return true;
else
return false;
}
void dfs(int x,int y)
{
map[x][y]='#';//更新地板
for(int i=0;i<4;i++)
{
int xx=x+dir[i][0];
int yy=y+dir[i][1];
if(limit(xx,yy)&&map[xx][yy]=='.')
{
sum++;
dfs(xx,yy);
}
}
}
int main()
{
while(scanf("%d %d",&m,&n)!=EOF)
{
if(n==0&&m==0)
break;
memset(map,0,sizeof(map));
for(int i=0;i<n;i++)
scanf("%s",map[i]);
sum=0;
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
if(map[i][j]=='@')
{
dfs(i,j);
break;
}
printf("%d\n",sum+1);//起点也是，所以+1
}
return 0;
}版权声明：本文为博主原创文章，未经博主允许不得转载。
标签：poj1979 poj 1979 red and black
原文地址：http://blog.csdn.net/su20145104009/article/details/46945355
