Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8911 Accepted Submission(s): 5535
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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0 此题中@也算一个可以走位置所以答案要加1，搜索时注意标记走过的点的位置以免重复计算 还有边界也要注意
#include<iostream>
#include<cstring>
using namespace std;
int ans,n,m;
int dy[4]={0,1,0,-1};
int dx[4]={1,0,-1,0};
char a[1005][1005],b[1005][1005];
void dfs(int x,int y)
{
int xx,yy,i;
for(i=0;i<4;i++)
{
xx=x+dx[i]; //此处注意不能直接用x+=d[x];否则回溯时不会回到x的初始值
yy=y+dy[i];
if(a[xx][yy]=='.'&&xx>=0&&xx<m&&yy>=0&&yy<n)
{
ans++;
a[xx][yy]='#';
dfs(xx,yy);
}
}
}
int main()
{
int i,j;
memset(b,0,sizeof(b));
while(cin>>n>>m,m&&n)
{
ans=0;
for(i=0;i<m;i++)
for(j=0;j<n;j++)
cin>>a[i][j];
for(i=0;i<m;i++)
for(j=0;j<n;j++)
if(a[i][j]=='@')
{a[i][j]='#';dfs(i,j);}
/*cout<<endl;
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
cout<<a[i][j];
cout<<endl;
}*/
cout<<ans+1<<endl;
}
return 0;
}
杭电 1312 red and black