Red and Black


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16081    Accepted Submission(s): 9912



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 ​​



题解：经典的搜索（bfs，dfs也可以）




TLE代码.....:搜得太深了。。。


#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<iomanip>
#include<algorithm>
typedef long long LL;
using namespace std;
int w,h;
char z[21][21];
int dfs(int i,int j){
    if(i<1||j>h||j<1||j>w)return 0;
    
    if(z[i][j]!='#')
    {
        z[i][j]='#';
        return 1+dfs(i-1,j)+dfs(i,j-1)+dfs(i+1,j)+dfs(i,j+1);  //TLE.... 
    }
    else return 0; 
}
int main()
{
    
     while(cin>>w>>h){
        if(w==0&&h==0)break;
        
        for(int i=1;i<=h;i++)
          for(int j=1;j<=w;j++)
            cin>>z[i][j];
            for(int i=1;i<=h;i++){
                for(int j=1;j<=w;j++)
                   if(z[i][j]=='@')
                   cout<<dfs(i,j)<<endl;
            }
     } 
    return 0;
}



AC代码：（BFS）


#include <stdio.h>
#include <queue>
#include <string.h>
#include <algorithm>
using namespace std;
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}}; //扩展的四个方向
char map[33][33];
int vis[33][33];
int n,m;
int sum;
struct node
{
  int x,y;
}
f[333];
#define inf 0xfffffff
void bfs(int x,int y)
{
  int i;
  queue<node>q;
  node st,ed;
  st.x=x;
  st.y=y;
  q.push(st);
  while(!q.empty())  //队列非空
  {
    st=q.front();
    q.pop();       
    for(i=0;i<4;i++)
    {
      ed.x=st.x+dir[i][0];
      ed.y=st.y+dir[i][1];
      if(ed.x>=n ||ed.y>=m ||ed.x<0 ||ed.y<0 ||map[ed.x][ed.y]=='#' ||map[ed.x][ed.y]=='@') //判断越界
        continue;
      map[ed.x][ed.y]='@';  //标志已经遍历过的
      sum++;
      q.push(ed);    
    }
  }
}
int main()
{
  int i,j,x,y;
  while(scanf("%d%d",&m,&n)!=EOF&&n!=0 &&m!=0)
  {
    for(i=0;i<n;i++)
      scanf("%s",map[i]);
    for(i=0;i<n;i++)
      for(j=0;j<m;j++)
      {
        if(map[i][j]=='@')  //寻找起始点
        {
          x=i;
          y=j;     
        }
      }
      sum=1;
      bfs(x,y);
      printf("%d\n",sum);
  }
  return 0;
}
//广搜


AC代码：（DFS）


#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
char map[33][33];
int vis[33][33];
int n,m;
int sum;
#define inf 0xfffffff
void dfs(int x,int y)
{
    int sx,sy,i;
    sum++;
    for(i=0;i<4;i++)
    {
        sx=x+dir[i][0];
        sy=y+dir[i][1];
        if(sx>=n ||sy>=m ||sx<0 ||sy<0 ||map[sx][sy]=='#' ||map[sx][sy]=='@')
            continue;
          map[sx][sy]='@';
              dfs(sx,sy);
    } 
}
int main()
{
    int i,j,x,y;
    while(scanf("%d%d",&m,&n)!=EOF&&n!=0 &&m!=0)
    {
        for(i=0;i<n;i++)
            scanf("%s",map[i]);
        for(i=0;i<n;i++)
            for(j=0;j<m;j++)
            {
                if(map[i][j]=='@')
                {
                    x=i;
                    y=j;         
                }
            }
            sum=0;
            dfs(x,y);
            printf("%d\n",sum);
    }
    return 0;
}
//深搜