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

 

 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;
#define maxn 25
#define INF 9999999
char mp[maxn][maxn];
struct Node{
    int x, y, t;
}s;
int W, H, sx, sy; //宽，高 
int dir[4][2] = {{0,1}, {1,0}, {-1,0},{0,-1}}; 
int timE[maxn][maxn], vis[maxn][maxn];
void BFS(Node start){
    queue <Node> Q;
    Q.push(start);
    vis[start.x][start.y] = 1;
    while(!Q.empty()){
        Node hd = Q.front(); Q.pop();
        for(int i = 0; i < 4; i++){
            int nx = hd.x + dir[i][0];
            int ny = hd.y + dir[i][1];
            if(nx >= 1 && nx <= H && ny >= 1 && ny <= W && mp[nx][ny] != '#' && !vis[nx][ny]){
                Node temp;
                temp.x = nx; temp.y = ny; temp.t = hd.t+1;
                vis[nx][ny] = 1;
                timE[nx][ny] = temp.t;
                    Q.push(temp);
                
                
            }
        }
    }

}
int main(){
    while(~scanf("%d%d", &W, &H)){
        if(W == 0 || H ==0) break;
        for(int i = 1; i <= H; i++){
            for(int j = 1; j <= W; j++){
                cin>>mp[i][j];
                if(mp[i][j] == '@'){
                    s.x = i;
                    s.y = j;
                    s.t = 0;
                }
                timE[i][j] = INF;
            }
        }
        timE[s.x][s.y] = 0;
        memset(vis, 0, sizeof(vis));
        BFS(s);
        int ans = 0;
        for(int i = 1; i <= H; i ++){
            for(int j = 1; j <= W; j++){
                if(timE[i][j] != INF)  ans++;
            }
        }
        printf("%d
", ans);
    }
    
    return 0;
}