HDU 1312 Red and Black (DFS入门)

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

4559613

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

 

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Eddy   |   We have carefully selected several similar problems for you:  1016 1242 1253 1240 1072 

题意：有红格子和黑格子。人只能走黑格子。数人能走多少个黑格子并输出。

转化后为数‘@’周围的‘.’的数量：

典型的DFS深搜........但是我一开始提交的时候老师CE，不知道什么原因，感觉代码很对，不知道什么原因，后来就想是不是用G++是原因，所以就用C++试了试，然后就AC了......

#include <iostream>#include <cstring>using namespace std;char str[25][25];int xx,yy;int w,h;int count;int dist[4][2]= {0,1,1,0,0,-1,-1,0};void dfs(int x,int y){    int sx,sy;    for(int i=0; i<4; i++)    {        sx=x+dist[i][0];        sy=y+dist[i][1];        if(sx>=1&&sx<=h&&sy>=1&&sy<=w&&str[sx][sy]=='.')        {            count++;            str[sx][sy]='#';            dfs(sx,sy);        }    }}int main(){    while(cin>>w>>h)    {        if(w==0&&h==0)            break;        for(int i=1; i<=h; i++)            for(int j=1; j<=w; j++)            {                cin>>str[i][j];                if(str[i][j]=='@')                {                    xx=i;                    yy=j;                }            }        count=1;        str[xx][yy]='#';        dfs(xx,yy);        cout<<count<<endl;    }    return 0;}