POJ——1979Red and Black

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 26448 Accepted: 14371

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

简单深搜

代码如下

#include<stdio.h>#include<string.h>#include<math.h>#include<stdlib.h>#include<ctype.h>#include<iostream>#include<string>#include<algorithm>#include<set>#include<vector>#include<queue>#include<map>#include<numeric>#include<stack>#include<list>const int INF=1<<30;const int inf=-(1<<30);const int MAX=100010;using namespace std;int n,m,t;char a[30][30];int s1[4]= {-1,1,0,0};int s2[4]= {0,0,-1,1};int visit[30][30];void dfs(int x,int y){    t++;    visit[x][y]=1;    for(int i=0; i<4; i++)    {        int t1=s1[i]+x;        int t2=s2[i]+y;        if(t1>=0&&t1<m&&t2>=0&&t2<n&&a[t1][t2]=='.'&&!visit[t1][t2])        {            visit[t1][t2]=1;            dfs(t1,t2);        }    }    return ;}int main(){    int q,p;    while(cin>>n>>m,n&&m)    {        memset(visit,0,sizeof(visit));        for(int i=0; i<m; i++)        {            for(int j=0; j<n; j++)            {                cin>>a[i][j];                if(a[i][j]=='@')                {                    p=i;                    q=j;                }            }        }        t=0;        dfs(p,q);        printf("%d\n",t);    }    return 0;}