Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9684 Accepted Submission(s): 6021
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
Sample Output
45
59
6
13
1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4 #include <string>
5 #include <iomanip>
6 #include <algorithm>
7 #include <queue>
8 #include <vector>
9 #include <map>
10 using namespace std;
11 #define maxn 25
12 #define INF 9999999
13 char mp[maxn][maxn];
14 struct Node{
15 int x, y, t;
16 }s;
17 int W, H, sx, sy; //宽，高
18 int dir[4][2] = {{0,1}, {1,0}, {-1,0},{0,-1}};
19 int timE[maxn][maxn], vis[maxn][maxn];
20 void BFS(Node start){
21 queue <Node> Q;
22 Q.push(start);
23 vis[start.x][start.y] = 1;
24 while(!Q.empty()){
25 Node hd = Q.front(); Q.pop();
26 for(int i = 0; i < 4; i++){
27 int nx = hd.x + dir[i][0];
28 int ny = hd.y + dir[i][1];
29 if(nx >= 1 && nx <= H && ny >= 1 && ny <= W && mp[nx][ny] != ‘#‘ && !vis[nx][ny]){
30 Node temp;
31 temp.x = nx; temp.y = ny; temp.t = hd.t+1;
32 vis[nx][ny] = 1;
33 timE[nx][ny] = temp.t;
34 Q.push(temp);
35
36
37 }
38 }
39 }
40
41 }
42 int main(){
43 while(~scanf("%d%d", &W, &H)){
44 if(W == 0 || H ==0) break;
45 for(int i = 1; i <= H; i++){
46 for(int j = 1; j <= W; j++){
47 cin>>mp[i][j];
48 if(mp[i][j] == ‘@‘){
49 s.x = i;
50 s.y = j;
51 s.t = 0;
52 }
53 timE[i][j] = INF;
54 }
55 }
56 timE[s.x][s.y] = 0;
57 memset(vis, 0, sizeof(vis));
58 BFS(s);
59 int ans = 0;
60 for(int i = 1; i <= H; i ++){
61 for(int j = 1; j <= W; j++){
62 if(timE[i][j] != INF) ans++;
63 }
64 }
65 printf("%d\n", ans);
66 }
67
68 return 0;
69 }
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