Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7337 Accepted Submission(s): 4591
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
广度搜索，需要用到队列：
//bfs
#include<stdio.h>
int map[][];
int queue[][];//队列
int sx[]={,,,-};
int sy[]={,,-,};
int h,w;
int ans;
void bfs(int x,int y)
{
int rear=,front=;
queue[rear][]=x;//enter queue
queue[rear][]=y;
rear++;
while(front<rear)
{
for(int t=;t<;t++)
{
x=queue[front][]+sx[t];
y=queue[front][]+sy[t];
if(map[x][y]=='.'&&x>=&&x<=w&&y>=&&y<=h)
{
queue[rear][]=x;
queue[rear][]=y;
map[x][y]='#';
rear++;
ans++;
}
}
front++;
} }
int main()
{
int i,j;
int x,y;
while(scanf("%d%d",&h,&w)!=EOF)
{
if(!h&&!w)break;
ans=;
for(i=;i<=w;i++)
{
getchar();
for(j=;j<=h;j++)
{
scanf("%c",&map[i][j]);
if(map[i][j]=='@')
x=i,y=j;
}
}
bfs(x,y);
printf("%d\n",++ans);
}
return ;
}
深度搜索：
#include<stdio.h>
int map[][];
int t,n,m;
int sx[]={,,-,};
int sy[]={-,,,};
void dfs(int h,int l)//深度搜索
{
int i,hx,hy;
t++;
map[h][l]='@';
for(i=;i<;i++)
{
hx=h+sx[i];
hy=l+sy[i];
if(hx>=&&hx<=m&&hy>=&&hy<=n&&map[hx][hy]=='.')
dfs(hx,hy);
}
}
int main()
{
int a,b,x,y;
while(scanf("%d%d",&n,&m)!=EOF)
{
if(n==&&m==)break;
for(a=;a<=m;a++)
{
getchar();
for(b=;b<=n;b++)
{
scanf("%c",&map[a][b]);
if(map[a][b]=='@')
{
x=a;
y=b;
}
}
}
t=;
dfs(x,y);
printf("%d\n",t);
}
return ;
}