DFS-HDU 1312 -Red and Black

D - Red and Black

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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 

 

这题已经不想在吐槽什么了，n,m 和正常的出相反 害我一些细节错 一直WA 还说到底还是自己太弱了

#include<iostream>using namespace std;#include<string.h>#define max 25char map[max][max];int mmin;long n,m,visited[max][max];long directions[4][2]={{0,-1},{0,1},{1,0},{-1,0}};void DFS(int x,int y){    int i,mx,my;    for(i=0;i<4;i++)        {            mx=x+directions[i][0];            my=y+directions[i][1];            if(mx>=0&&mx<m&&my>=0&&my<n)                    {                        if(!visited[mx][my]&&map[mx][my]=='.')                        {                        visited[mx][my]=1;                        mmin++;                           DFS(mx,my);                        }                    }        }}int main(){    int i,j,a,b;    while(cin>>n>>m)    {    if(n==0&&m==0)break;        memset(visited,0,sizeof(visited));        for(i=0;i<m;i++)        {            for(j=0;j<n;j++)            {                cin>>map[i][j];                if(map[i][j]=='@')                    {                       a=i,b=j;                        }            }        }        mmin=1;        visited[a][b]=1;        DFS(a,b);        cout<<mmin<<endl;    }    return 0;}