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  • hdu-1312

    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

    AC代码为:

    #include<cstdio>#include<iostream>#include<cmath>#include<cstring>using namespace std;int sum,H,W;int dx[4]={1,0,-1,0};int dy[4]={0,1,0,-1};char str[21][21];int vis[21][21];void f(int xl,int yl){vis[yl][xl]=1;sum++;for(int i=0;i<4;i++){int cx=xl+dx[i];int cy=yl+dy[i];if(cx>=0 && cx<W && cy>=0 && cy<H && !vis[cy][cx] && str[cy][cx]!='#')f(cx,cy);}}int main(){int x,y;while(~scanf("%d%d",&W,&H)){if(H==0 || W==0)break;for(int i=0;i<H;i++){scanf("%s",str[i]);}for(int i=0;i<H;i++){for(int j=0;j<W;j++)if(str[i][j]=='@'){x=j;y=i;}}sum=0;memset(vis,0,sizeof(vis));f(x,y);printf("%d ",sum);}return 0;}




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  • 原文地址:https://www.cnblogs.com/csushl/p/9386592.html
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