Red and Black

最新推荐文章于 2022-01-12 23:41:22 发布

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本文介绍了一个使用深度优先搜索（DFS）解决的问题，即在一个由黑色和红色瓷砖组成的矩形房间里，从一个黑色瓷砖出发，只能在黑色瓷砖上行走，计算可以到达的黑色瓷砖数量。通过输入房间的宽度和高度，以及房间布局，输出可以到达的黑色瓷砖总数。

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.

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)

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

分析：此题是简单的搜索题，本人用的是深搜dfs

代码：

#include<stdio.h>
#include<string.h>
char a[130][130];
int vis[130][130];
int n,m,count;
//int x[4]={1,-1,0,0};
//int y[4]={0,0,-1,1};
void dfs(int r,int t,int s)
{
int i;
if(r>=m||r<0||t>=n||t<0||vis[r][t])
return ;
if(a[r][t]=='#')
return ;
++count;
vis[r][t]=1;
dfs(r,t+1,s);
dfs(r-1,t,s);
dfs(r,t-1,s);
dfs(r+1,t,s);
//vis[r][t]=0;
// for( i=0;i<4;i++)
// {
// dfs(x[i]+r,y[i]+t,s);
// }
}
int main(){
int i,j,k,r,t;
while(~scanf("%d%d",&n,&m))
{
for(i=0;i<m;i++)
scanf("%s",a[i]);
memset(vis,0,sizeof(vis));
for(i=0;i<m;i++)
for(j=0;j<n;j++)
if(a[i][j]=='@')
{
r=i;
t=j;
//break;
}
count=0;
dfs(r,t,k);
printf("%d\n",count);
}
return 0;
}