HDU 1312 Red and Black 第一题搜索！

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12935    Accepted Submission(s): 8006

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

 

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这是我写搜索的第一题，dfs bfs都是在这一题上学会的。从今天开始我要开始搞搜索了，我必须全面提高自己的能力，所有的类型都得有个差不多，不然根本没法和他们讨论。

 

bfs & dfs

 #include<math.h>
 #include<stdio.h>
 #include<queue>
 #include<string.h>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 #define N 1234
 struct point
 {
     int x,y;
 }st;

 int n,m,ans;
 int dx[]={,,,-};
 int dy[]={,-,,};
 char mat[N][N];
 int  vis[N][N];

 void bfs()
 {
     queue<point>q;
     q.push(st);vis[st.x][st.y]=;
     while(!q.empty())
     {
         point cur=q.front();
         q.pop();
         ans++;
         for(int i=;i<;i++)
         {
             point next=cur;
             next.x+=dx[i];next.y+=dy[i];
             if(next.x<||next.x>n||next.y<||next.y>m)continue;
             if(mat[next.x][next.y]=='#'||vis[next.x][next.y]==)continue;
             q.push(next);vis[next.x][next.y]=;
         }
     }
 }
 void dfs(int x,int y)
 {
     if(x<||x>n||y<||y>m)return;
     if(mat[x][y]=='#'||vis[x][y]==)return;
     vis[x][y]=;
     ans++;
     for(int i=;i<;i++)
         dfs(x+dx[i],y+dy[i]);
 }

 int main()
 {
     while(~scanf("%d%d",&m,&n)&&(m+n))
     {
         memset(vis,,sizeof(vis));
         for(int i=;i<=n;i++)
             for(int j=;j<=m;j++)
             {
                 scanf(" %c",&mat[i][j]);
                 if(mat[i][j]=='@')
                 {
                     st.x=i;st.y=j;
                 }
             }
             ans=;
             bfs();
 //            dfs(st.x,st.y);
             cout<<ans<<endl;
     }
     return ;
 }