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  • Red and Black HDU

    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. 

    InputThe 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) 
    OutputFor 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

    思路:BFS
    AC Code:
    #include<iostream>
    #include<cstdio>
    #include<queue>
    #include<utility>
    #include<cstring>
    using namespace std;
    typedef pair<int,int> P;
    char maps[25][25];
    int vis[25][25];
    int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
    int W,H,cnt;
    void bfs(P p){
        queue<P> q;q.push(p);
        while(!q.empty() ){
            p=q.front() ;q.pop() ;
            int x=p.first,y=p.second;
            for(int i=0;i<4;i++)
            {
                int nx=x+dx[i],ny=y+dy[i];
                if(nx>=0&&nx<H&&ny>=0&&ny<W&&maps[nx][ny]!='#'&&vis[nx][ny]==0){
                    cnt++;
                    vis[nx][ny]=1;
                    q.push(P(nx,ny));  
                }
            }
        }     
        
    }
    int main(){
        while(~scanf("%d%d",&W,&H)){
            if(W==0&&H==0) break;
            int x,y;
            getchar();
            memset(vis,0,sizeof(vis));
            for(int i=0;i<H;i++){
                for(int j=0;j<W;j++){
                    scanf("%c",&maps[i][j]);
                    if(maps[i][j]=='@') { x=i; y=j; }    
                }
                getchar();
            }
            cnt=1;vis[x][y]=1;
            bfs(P(x,y));
            printf("%d
    ",cnt);
        }
    }
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  • 原文地址:https://www.cnblogs.com/astonc/p/9900578.html
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