ACM-搜索之Red and Black——hdu1312

Red and Black
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

很简单的一道题目，我觉得深搜广搜应该都能过= =。就是从@开始所有向四个方向走，查看。的数量，当然

走到边界返回，碰到#返回。

这道题目有一点比较坑，就是输入的两个数，第一个代表的是列数第二个是行数，我就因为这个纠结了会，

后来调试才发现这个问题，o(╯□╰)o囧啊。。。。

DFS：

#include <iostream>
#include <string.h>
using namespace std;

// mapp存储地图，dis数组存储四个方向，vis存储是否走过
char mapp[21][21];
int n,m,num,vis[21][21],dis[4][2]={-1,0,0,1,0,-1,1,0};

// 判断是否越界，或者是否碰到#，或者是否已经遍历过
bool judge(int x,int y)
{
    if(x<0 || y<0 ||x>=m || y>=n)   return 0;
    if(mapp[x][y]=='#' || vis[x][y]==1)    return 0;
    return 1;
}

void dfs(int x,int y)
{
    int i,xx,yy;
    for(i=0;i<4;++i)
    {
        xx=x+dis[i][0];
        yy=y+dis[i][1];
        if(judge(xx,yy))
        {
            vis[xx][yy]=1;
            num+=1;
            dfs(xx,yy);
        }
    }
}

int main()
{
    int i,j,s_x,s_y;
    while(cin>>n>>m)
    {
        if(n==0 && m==0)    break;

        // 输入地图，记录起始点
        for(i=0;i<m;++i)
            for(j=0;j<n;++j)
            {
                cin>>mapp[i][j];
                if(mapp[i][j]=='@'){s_x=i;s_y=j;}
            }

        memset(vis,0,sizeof(vis));

        // 从第一个开始走，第一步要算上
        vis[s_x][s_y]=1;
        num=1;
        dfs(s_x,s_y);
        cout<<num<<endl;


    }
    return 0;
}

BFS：

#include <iostream>
using namespace std;

// mapp存储地图，dis数组存储四个方向，vis存储是否走过
char mapp[21][21];
int n,m;

int bfs(int x,int y)
{
    if(x<0 || y<0 ||x>=m || y>=n)   return 0;
    if(mapp[x][y]=='#')    return 0;
    mapp[x][y]='#';

    return bfs(x-1,y)+bfs(x+1,y)+bfs(x,y-1)+bfs(x,y+1)+1;
}

int main()
{
    int i,j,s_x,s_y;
    while(cin>>n>>m)
    {
        if(n==0 && m==0)    break;

        // 输入地图，记录起始点
        for(i=0;i<m;++i)
            for(j=0;j<n;++j)
            {
                cin>>mapp[i][j];
                if(mapp[i][j]=='@'){s_x=i;s_y=j;}
            }

        int num=bfs(s_x,s_y);
        cout<<num<<endl;
    }
    return 0;
}