HDU 1312 Red and Black(最简单也是最经典的搜索)
传送门:
http://acm.hdu.edu.cn/showproblem.php?pid=1312
Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25397 Accepted Submission(s): 15306
Write a program to count the number of black tiles which he can reach by repeating the moves described above.
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)
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0
59
6
13
#include<bits/stdc++.h>
using namespace std;
#define max_v 25
char G[max_v][max_v];
int n,m;
int sx,sy;
int step;
int dir[][]={,,,,,-,-,};
void dfs(int x,int y)
{
int xx,yy;
for(int i=;i<;i++)
{
xx=x+dir[i][];
yy=y+dir[i][];
if(xx>=&&xx<n&&yy>=&&yy<m&&G[xx][yy]!='#')
{
step++;
G[xx][yy]='#';
dfs(xx,yy);
}
}
}
int main()
{
while(~scanf("%d %d",&m,&n))
{
if(n==&&m==)
break;
getchar();
for(int i=;i<n;i++)
{
for(int j=;j<m;j++)
{
cin>>G[i][j];
if(G[i][j]=='@')
{
sx=i;
sy=j;
}
}
}
step=;
G[sx][sy]='#';
dfs(sx,sy);
cout<<step<<endl;
}
return ;
}
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