hdu1312 “Red and Black”——DFS与递归的应用

Red and Black
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).

分析:
除了BFS以外,还可以使用DFS,设num=1,从起点开始,每次符合就num++,实在无法找到再返回,这符合递归思想。

代码如下:

#include<bits/stdc++.h>
using namespace std;
int W,H,num;
char room[25][25];
#define CHECK(x,y)(x<W&&x>=0&&y<H&&y>=0)
int dir[4][2]={
    {-1,0},{0,-1},{1,0},{0,1}
};
void dfs(int dx,int dy){
    room[dx][dy]='#';
    num++;
    for(int i=0;i<4;i++){
        int newx=dx+dir[i][0];
        int newy=dy+dir[i][1];
        if(CHECK(newx,newy)&&room[newx][newy]=='.')
            dfs(newx,newy);
    }
}
int main(){
    int x,y,dx,dy;
    while(~scanf("%d%d",&W,&H)){
        if(!W&&!H)break;
        for(y=0;y<H;y++){
            for(x=0;x<W;x++){
                cin>>room[x][y];
                if(room[x][y]=='@')dx=x,dy=y;
            }
        }
        num=0;
        dfs(dx,dy);
        printf("%d\n",num);
    }
}

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