HDU 1312Red and Black(简单搜索 bfs或dfs)

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6908    Accepted Submission(s): 4387

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

 

题目意思就不多说了，很容易理解，@为起点，可以上下左右四个方向扩展，

只能走点，问可以到达多少个位置。

DFS AC代码：

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
using namespace std;
int r,c;
int dir[4][2]={{0,-1},{0,1},{1,0},{-1,0}};
char s[25][25];
int visi[25][25];

void dfs(int a,int b)
{
    int i;
    for(i=0;i<4;i++)
    {
        int ta,tb;
        ta=a+dir[i][0],tb=b+dir[i][1];
        if(ta>=0&&ta<r&&tb>=0&&tb<c)
        {
            if(!visi[ta][tb]&&s[ta][tb]=='.')
            {
                visi[ta][tb]=1;
                dfs(ta,tb);
            }
        }
    }
}

int main()
{
    int i,j;
    while(scanf("%d%d",&c,&r))
    {
        if(!r&&!c) break;
        memset(visi,0,sizeof(visi));
        for(i=0;i<r;i++)
            scanf("%s",s[i]);
        int a,b;  //把起点的位置找到
        for(i=0;i<r;i++)
            for(j=0;j<c;j++)
                if(s[i][j]=='@')
                {
                    a=i,b=j;
                    break;
                }
        visi[a][b]=1;
        dfs(a,b);

        int res=0;  //统计访问的个数
        for(i=0;i<r;i++)
            for(j=0;j<c;j++)
                if(visi[i][j]==1)
                    res++;
        printf("%d\n",res);
    }
    return 0;
}

//0MS

BFS AC代码：

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<queue>
using namespace std;
int r,c;
int dir[4][2]={{0,-1},{0,1},{1,0},{-1,0}};
char s[25][25];
int visi[25][25];

void bfs(int t)
{
    queue<int> mq;
    mq.push(t);
    while(!mq.empty())
    {
        int p=mq.front();
        mq.pop();
        int ta=p/c,tb=p%c;
        for(int i=0;i<4;i++)
        {
            int tta=dir[i][0]+ta,ttb=dir[i][1]+tb;
            if(tta>=0&&tta<r&&ttb>=0&&ttb<c)
            {
                if(!visi[tta][ttb]&&s[tta][ttb]=='.')
                {
                    visi[tta][ttb]=1;
                    mq.push(tta*c+ttb);
                }
            }
        }
    }
}

int main()
{
    int i,j;
    while(scanf("%d%d",&c,&r))
    {
        if(!r&&!c) break;
        memset(visi,0,sizeof(visi));
        for(i=0;i<r;i++)
            scanf("%s",s[i]);
        int a,b;  //把起点的位置找到
        for(i=0;i<r;i++)
            for(j=0;j<c;j++)
                if(s[i][j]=='@')
                {
                    a=i,b=j;
                    break;
                }
        visi[a][b]=1;
        bfs(a*c+b);

        int res=0;  //统计访问的个数
        for(i=0;i<r;i++)
            for(j=0;j<c;j++)
                if(visi[i][j]==1)
                    res++;
        printf("%d\n",res);
    }
    return 0;
}

//15MS