dfs学习总结

今天做到了dfs的训练，感觉和bfs有相似之处，接下来用一道题来总结一下方法，可类比bfs。

上题：

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

 4559613 

 

解题代码：

#include<stdio.h>#include<string.h>char point[25][25];bool flag[25][25];              //flag对应着我们需要研究的点point，用来标记是否曾经到过int dx[4]={1,-1,0,0};int dy[4]={0,0,-1,1};int count;void dfs(int x0,int y0,int r,int c){    for(int i=0;i<4;i++)        //for循环用来探索所有邻接点    {        int tempx=x0+dx[i],tempy=y0+dy[i];        if(tempx<c&&tempx>=0&&tempy<r&&tempy>=0&&flag[tempy][tempx]==false&&point[tempy][tempx]=='.')        {            count++;            flag[tempy][tempx]=true;        //满足条件且未标记的标记上            dfs(tempx,tempy,r,c);           //通过递归来实现顺藤摸瓜的效果        }    }    return ;}void make_set(){    for(int i=0;i<25;i++)        for(int j=0;j<25;j++)            flag[i][j]=false;    return ;}int main(){    int c,r,x0,y0;    while(1)    {        count=1;        scanf("%d%d",&c,&r);        getchar();        if(c==0&&r==0)            break;        for(int i=0;i<r;i++)        {            for(int j=0;j<c;j++)            {                point[i][j]=getchar();                if(point[i][j]=='@')                {                    x0=j;                    y0=i;                }            }            getchar();        }        make_set();        dfs(x0,y0,r,c);        printf("%d\n",count);    }    return 0;}

与bfs的区别：bfs是通过队列来进行逐层探索，而dfs则是沿着一个路径探索下去一直到底，到底后再返回沿其他路径探索，就和顺藤摸瓜差不多，大体上两者达到的效果基本一致（特殊情况效果有区别），两者均有缺点，bfs在编写代码时较麻烦，用到队列，一般要使用结构体，而dfs则是由于其使用递归调用，大数据易超时。

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赶紧跑回来加一句：最短路径用bfs！！