POJ1979——Red and Black

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 31223 Accepted: 17049

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

题意：一个w*h的矩阵，从‘@’出发，如果是 ' . ' 则可以通行，如果是' # ' 则不可通行。计算一条路线的最大距离，加上本身所在的位置。

解：使用dfs进行搜索，并使用v[i][j]来标记是否走过此处，如果走过则标记为1。进行上下左右四个方向的搜索。

#include<stdio.h>#include<string.h>char a[25][25];int v[25][25];int n,t,count;void dfs(int x,int y){if(x<0 || y<0 || x>=n || y>=t) return ;if(a[x][y]=='#')return;if(v[x][y])return;else if(a[x][y]=='.'){v[x][y]=1;count++;}dfs(x+1,y);dfs(x-1,y);dfs(x,y+1);dfs(x,y-1);}int main(){int ii,jj;while(scanf("%d%d",&t,&n) && n && t){getchar();for(int i=0;i<n;i++){for(int j=0;j<t;j++){scanf("%c",&a[i][j]);if(a[i][j]=='@'){ii=i;jj=j;}}getchar();}memset(v,0,sizeof(v));count=0;dfs(ii,jj);printf("%d\n",count+1);}return 0;}