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

 

Recommend

Eddy   |   We have carefully selected several similar problems for you:  1241 1016 1010 1372 1242

 

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[]={0,0,1,-1};
int dy[]={1,-1,0,0};
char mat[N][N];
int  vis[N][N];

void bfs()
{
    queue<point>q;
    q.push(st);vis[st.x][st.y]=1;
    while(!q.empty())
    {
        point cur=q.front();
        q.pop();
        ans++;
        for(int i=0;i<4;i++)
        {
            point next=cur;
            next.x+=dx[i];next.y+=dy[i];
            if(next.x<1||next.x>n||next.y<1||next.y>m)continue;
            if(mat[next.x][next.y]=='#'||vis[next.x][next.y]==1)continue;
            q.push(next);vis[next.x][next.y]=1;
        }
    }
}
void dfs(int x,int y)
{
    if(x<1||x>n||y<1||y>m)return;
    if(mat[x][y]=='#'||vis[x][y]==1)return;
    vis[x][y]=1;
    ans++;
    for(int i=0;i<4;i++)
        dfs(x+dx[i],y+dy[i]);
}

int main()
{
    while(~scanf("%d%d",&m,&n)&&(m+n))
    {
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
            {
                scanf(" %c",&mat[i][j]);
                if(mat[i][j]=='@')
                {
                    st.x=i;st.y=j;
                }
            }
            ans=0;
            bfs();
//            dfs(st.x,st.y);
            cout<<ans<<endl;
    }
    return 0;
}