HDU1312 Red and Black

                                                      Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6103    Accepted Submission(s): 3887

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
   
   
   
   

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

 

Recommend

Eddy

解题思路：简单搜索题，深度优先搜索，广度优先搜索都可以。只要做到两点，题目即可AC。1，把所有能到达的地方走一遍，2，同一点计数不能有重复，走过后标记即可。

#include<cstdio> #include<queue> #include<string.h> using namespace std; char map[20][20]; int flag[20][20]; int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}}; int n,m; int sx,sy,maxstep=1; struct Node {     int x;     int y; }; void bfs() {     queue<Node> Q;     int x,y;     int k;     Node first,next;     first.x=sx;     first.y=sy;     flag[sx][sy]=1;     Q.push(first);     while(!Q.empty())     {         first=Q.front();         Q.pop();         for(k=0;k<4;k++)         {             x=first.x+dir[k][0];             y=first.y+dir[k][1];             if(x<0||x>=n||y<0||y>=m)                 continue;             if(map[x][y]=='.'&&flag[x][y]==0)             {                  next.x=x;                 next.y=y;                 maxstep++;                 Q.push(next);                 flag[x][y]=1;    //记得标记啊             }         }     } } int main() {     int i,j;     while(scanf("%d%d",&m,&n)&&n||m)     {         memset(flag,0,sizeof(flag));         maxstep=1;         for(i=0;i<n;i++)         {             scanf("%s",map[i]);             for(j=0;j<m;j++)                 if(map[i][j]=='@')                 {                     sx=i;                     sy=j;                 }         }         bfs();         printf("%d\n",maxstep);     }     return 0; }