hdu 1312 Red and Black（BFS水题）

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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 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
", ans);
66     }
67     
68     return 0;
69 }