POJ 1979 Red and Black DFS水题

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 31119 Accepted: 16977

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：输入地图，@为起点，.可以走，#不可以走，求出可以走到的区域点数，包括@。

    分析：DFS水题。设置全局变量res时注意要清零。见AC代码：

#include<stdio.h>#include<string.h>const int maxn=21;char map[maxn][maxn];int  vis[maxn][maxn];int res=0;int w,h;int dx[4]= {0,0,-1,1};int dy[4]= {-1,1,0,0};void dfs(int x,int y){if(x<0||x>=h||y<0||y>=w)return;if(!vis[x][y]&&(map[x][y]=='.'||map[x][y]=='@')){res++;vis[x][y]=1;//printf("++\n");for(int i=0; i<4; i++)dfs(x+dx[i],y+dy[i]);}}int main(){while(~scanf("%d%d",&w,&h)&&w&&h){res=0;memset(vis,0,sizeof(vis));for(int i=0; i<h; i++)scanf("%s",map[i]);int xs,ys;for(int i=0; i<h; i++)for(int j=0; j<w; j++)if(map[i][j]=='@'){xs=i;ys=j;break;}//printf("%d %d \n",xs,ys);dfs(xs,ys);printf("%d\n",res);}}

    刷题时需细心仔细，避免无谓错误。DFS需要多刷。

    特记下，以备后日回顾。