Red and Black
Tme Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18126 Accepted Submission(s): 11045
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.
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)
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).
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
题意就是 找从@点开始出发，问所能到达的'.'点的个数（最开始理解错了意思，以为是一条路径的最大值，后面
才发现是可以到达的‘.'点 ），直接DFS，
#include<iostream>
#include<cstdio>
#include<cmath>
#include<map>
#include<cstdlib>
#include<vector>
#include<set>
#include<queue>
#include<cstring>
#include<string.h>
#include<algorithm>
#define INF 0x3f3f3f3f
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const int N = 1e6+10;
const ll mod = 1e9+7;
int t=0,flag=0;
char s[26][26];
int visited[26][26];
int m,n;
int _x[4]={0,1,-1,0};
int _y[4]={1,0,0,-1};
int ans;
void DFS(int x,int y){
for(int t=0;t<4;t++){
int i=x+_x[t];
int j=y+_y[t];
if(i>=0&&j>=0&&visited[i][j]==0&&s[i][j]=='.'&&i<=n-1&&j<=m-1)
{
ans++;
visited[i][j]=1;
DFS(i,j);
}
}
}
int main(){
while(scanf("%d%d",&m,&n)!=EOF){
if(m==0&&n==0)break;
memset(visited,0,sizeof(visited));
int ii,jj;
int i,j;
for(ii=0;ii<n;ii++){
for(jj=0;jj<m;jj++){
cin>>s[ii][jj];
if(s[ii][jj]=='@'){i=ii;j=jj;}
}
}
ans=1;
visited[i][j]=1;
DFS(i,j);
cout<<ans<<endl;
}
}