Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8435 Accepted Submission(s): 5248
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
....#.
.....#
......
......
......
......
......
#@...#
.#..#. .#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
........... ..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#.. ..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
Sample Output
45
59
6
13
Source
Asia 2004, Ehime (Japan), Japan Domestic
Recommend
Eddy | We have carefully selected several similar problems for you: 1372 1242 1240 1072 1258
DFS搜索，入门题。
规定地图中有可通行的位置，也有不可通行的位置，已知起点，求一个连通分量。说白了就是求一个点的与它相连的部分。在这道题里输出相连的位置的数目。
思路是从起点开始，遍历每一个到达的点的四个方向，到达一个位置就将这个位置的字符变成不可走的'#'，并且计数+1。其实就是计数将可走变成不可走的操作进行了多少次。
代码一：
#include <iostream>
using namespace std;
int cnt;
char a[][];
int n,m;
int dx[] = {,,,-}; //方向
int dy[] = {,,-,};
bool judge(int x,int y)
{
if(x< || x>n || y< || y>m)
return ;
if(a[x][y]=='#')
return ;
return ;
}
void dfs(int cx,int cy)
{
cnt++;
a[cx][cy] = '#';
int i;
for(i=;i<;i++){
int nx = cx + dx[i];
int ny = cy + dy[i];
if(judge(nx,ny))
continue;
//可以走
dfs(nx,ny);
}
}
int main()
{
while(cin>>m>>n){
if(n== && m==) break;
int i,j;
int x,y;
for(i=;i<=n;i++)
for(j=;j<=m;j++){
cin>>a[i][j];
if(a[i][j]=='@') //记录开始的位置
x=i,y=j;
}
cnt = ;
dfs(x,y);
cout<<cnt<<endl;
}
return ;
}
这种方法直接返回结果，两种不同的写法，代码二：
#include <iostream>
using namespace std;
char a[][];
int n,m;
int dx[] = {,,,-}; //方向
int dy[] = {,,-,};
bool judge(int x,int y)
{
if(x< || x>n || y< || y>m)
return ;
if(a[x][y]=='#')
return ;
return ;
}
int dfs(int cx,int cy)
{
int i,sum=;
a[cx][cy] = '#'; //走过的这一步覆盖
for(i=;i<;i++){
int nx = cx + dx[i];
int ny = cy + dy[i];
if(judge(nx,ny))
continue;
//可以走
sum+=dfs(nx,ny);
}
return sum==?:sum+;
}
int main()
{
while(cin>>m>>n){
if(n== && m==) break;
int i,j;
int x,y;
for(i=;i<=n;i++)
for(j=;j<=m;j++){
cin>>a[i][j];
if(a[i][j]=='@') //记录开始的位置
x=i,y=j;
}
cout<<dfs(x,y)<<endl;
}
return ;
}
Freecode : www.cnblogs.com/yym2013
