POJ1979 Red and Black(深搜DFS)

题目：

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 32833 Accepted: 17852

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) 
The end of the input is indicated by a line consisting of two zeros. 

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

Source

Japan 2004 Domestic

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思路：

题意很简单,"#"代表墙，"."可以走，"@"代表出发点，问最多能走多少步（不走重复）,简单深搜，直接放代码

代码：

#include <stdio.h>#include <string.h>#include <iostream>#include <stack>#include <queue>#include <vector>#include <algorithm>#define mem(a,b) memset(a,b,sizeof(a))using namespace std;int w,h,x1,y1,maxx;char map[50][50];int vis[50][50];int go[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};void dfs(int x,int y){    for(int i=0; i<4; i++)    {        int xx=x+go[i][0];        int yy=y+go[i][1];        if(xx>=0&&xx<h&&yy>=0&&yy<w&&vis[xx][yy]==0&&map[xx][yy]=='.')        {            vis[xx][yy]=1;            dfs(xx,yy);            maxx++;        }    }}int main(){    while(~scanf("%d%d",&w,&h)&&(w||h))    {        maxx=1;        mem(vis,0);        for(int i=0; i<h; i++)        {            scanf("%s",map[i]);            for(int j=0; j<w; j++)                if(map[i][j]=='@')                {                    x1=i;                    y1=j;                }        }        dfs(x1,y1);        printf("%d\n",maxx);    }    return 0;}