【HDU 1312 Red and Black】+ DFS + BFS

Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 20903 Accepted Submission(s): 12726

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

45

59

6

13

裸 dfs || bfs

附上 dfs + bfs

AC代码：

DFS ：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX = 22;
typedef long long LL;
char s[MAX][MAX];
int vis[MAX][MAX],n,m,ans;
int fx[4] = {0,0,-1,1},fy[4] = {-1,1,0,0};
void dfs(int x,int y){
vis[x][y] = 1,ans++;
for(int i = 0; i < 4; i++){
int xx = x + fx[i],yy = y + fy[i];
if(xx >= 0 && yy >= 0 && xx < n && yy < m && !vis[xx][yy] && s[xx][yy] == '.')
dfs(xx,yy);
}
}
int main()
{
while(~scanf("%d %d",&m,&n),n){
int x,y;
memset(vis,0,sizeof(vis));
for(int i = 0; i < n; i++)
scanf("%s",s[i]);
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
if(s[i][j] == '@')
x = i,y = j;
ans = 0;
dfs(x,y);
printf("%d\n",ans);
}
return 0;
}

bfs :

#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX = 22;
typedef long long LL;
char s[MAX][MAX];
int vis[MAX][MAX],n,m,ans;
int fx[4] = {0,0,-1,1},fy[4] = {-1,1,0,0};
struct node{
int x,y;
};
void bfs(int x,int y){
vis[x][y] = 1;
node o;
o.x = x,o.y = y;
queue <node> q;
q.push(o);
while(!q.empty()){
o = q.front();
ans++,q.pop();
for(int i = 0; i < 4; i++){
int xx = o.x + fx[i],yy = o.y + fy[i];
if(xx >= 0 && yy >= 0 && xx < n && yy < m && !vis[xx][yy] && s[xx][yy] == '.'){
node w;
vis[xx][yy] = 1;
w.x = xx,w.y = yy;
q.push(w);
}
}
}
}
int main()
{
while(~scanf("%d %d",&m,&n),n){
int x,y;
memset(vis,0,sizeof(vis));
for(int i = 0; i < n; i++)
scanf("%s",s[i]);
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
if(s[i][j] == '@')
x = i,y = j;
ans = 0;
bfs(x,y);
printf("%d\n",ans);
}
return 0;
}