dfs 队列

题目来源  poj 1562

Description

The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid.

Input

The input contains one or more grids. Each grid begins with a line containing m and n, the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise 1 <= m <= 100 and 1 <= n <= 100. Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either `*', representing the absence of oil, or `@', representing an oil pocket.

Output

are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.

Sample Input

1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5
****@
*@@*@
*@**@
@@@*@
@@**@
0 0

Sample Output

0
1
2
2

ac代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define Swap(a,b,t) t=a,a=b,b=t
#define Mem0(x) memset(x,0,sizeof(x))
#define Mem1(x) memset(x,-1,sizeof(x))
#define MemX(x) memset(x,0x3f,sizeof(x));
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f;
const double eps=1e-12;
struct s{
int x,y;
char c;
}map[110][110];
const int dir[8][2]={{1,0},{1,-1},{1,1},{0,1},{0,-1},{-1,-1},{-1,0},{-1,1}};
queue<s> q;
int n,m,ans;
void dfs(int x,int y)
{
s p,next;
while (!q.empty()){
p=q.front();
q.pop();
for (int k=0;k<8;k++){
next.x=p.x+dir[k][0];
next.y=p.y+dir[k][1];
if (map[next.x][next.y].c=='@'){
q.push(map[next.x][next.y]);
map[next.x][next.y].c='*';
}
}
}
return ;
}
int main()
{
while (cin>>n>>m&&n&&m){
ans=0;
while (!q.empty())
q.pop();
for (int i=1;i<=n;i++){
for (int j=1;j<=m;j++){
cin>>map[i][j].c;
if (map[i][j].c=='@'){
map[i][j].x=i;
map[i][j].y=j;
}
}
}
for (int i=1;i<=n;i++){
for (int j=1;j<=m;j++){
if (map[i][j].c=='@'){
q.push(map[i][j]);
map[i][j].c='*';
ans++;
dfs(i,j);
}
}
}
printf("%d\n",ans);
}
return 0;
}

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26317    Accepted Submission(s): 15909

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

 

ac代码：

#include<iostream> //hdu   1312
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define Swap(a,b,t) t=a,a=b,b=t
#define Mem0(x) memset(x,0,sizeof(x))
#define Mem1(x) memset(x,-1,sizeof(x))
#define MemX(x) memset(x,0x3f,sizeof(x));
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f;
const double eps=1e-12;
int n,m,ans;
char temp[25];
const int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
struct s{
int x,y;
char c;
}map[25][25];
queue <s> q;
void dfs(int x,int y)
{
s p,next;
while (!q.empty()){
p=q.front();
q.pop();
for (int k=0;k<4;k++){
next.x=p.x+dir[k][0];
next.y=p.y+dir[k][1];
if (map[next.x][next.y].c=='.'&&next.x<=n&&next.y<=m){
ans++;
q.push(map[next.x][next.y]);
map[next.x][next.y].c='#';
}
}
}
return ;
}
int main()
{
while (cin>>m>>n&&n&&m){
while (!q.empty())
q.pop();
int start_x,start_y;
ans=1;
for(int i=1;i<=n;i++){
for (int j=1;j<=m;j++){
cin>>map[i][j].c;
if (map[i][j].c=='.'){
map[i][j].x=i;
map[i][j].y=j;
}
else if (map[i][j].c=='@'){
map[i][j].x=start_x=i;
map[i][j].y=start_y=j;
q.push(map[i][j]);
}
}
}
dfs(start_x,start_y);
printf("%d\n",ans);
}
return 0;
}