Red and Black 深搜



B - Red and Black

Crawling in process...Crawling failedTime Limit:1000MS    Memory Limit:30000KB     64bit IO Format:%I64d & %I64u

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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

题意：就是数一下一个@所在的位置可以走多少步的问题。

思路：找到@的地方进行深搜有四个方向，没有剪枝

AC代码：

#include<iostream>#include<cstdio>using namespace std;

char map[20][20];int b[4]={0,1,-1,0};int bb[4]={1,0,0,-1};int r,c;int sx,sy;int tol;int ok(int x,int y){    if(x<0||y<0||x>=r||y>=c||map[x][y]=='#')return 0;    return 1;}void dfs(int x,int y){    for(int i=0;i<4;i++){        map[x][y]='#';        if(ok(x+b[i],y+bb[i])){            tol++;            dfs(x+b[i],y+bb[i]);        }    }}int main (){    while(scanf("%d%d",&c,&r)!=EOF&&r!=0&&c!=0){        getchar();        for(int i=0;i<r;++i){                for(int j=0;j<c;j++){                   scanf("%c",&map[i][j]);//记住此处gets（）是不行的，可能输入空格                   if(map[i][j]=='@'){                    sx=i;                    sy=j;                   }                }        if(i<r-1){getchar();}

        }        tol=1;        dfs(sx,sy);        printf("%d\n",tol);    }    return 0;}

这倒以前做过的，看见还是发现有点陌生，长时间不敲就忘了怎么弄，好好强化。。