POJ 1979-Red and Black【基础DFS】

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 27684 Accepted: 15033

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

解题思路

找到@可以行动的区域，看清楚N和M谁代表行谁代表列。

#include<stdio.h>#include<string.h>#include<algorithm>#define INF 0x3f3f3f3f#define M 300using namespace std;int dx[4]={0,1,-1,0};int dy[4]={1,0,0,-1};int n,m,x,y; bool vis[M][M];char map[M][M];int ans;void F(int xx,int yy){int i,j;for(i=0;i<4;i++){if(map[xx+dx[i]][yy+dy[i]]=='.'&&vis[xx+dx[i]][yy+dy[i]]==false&&xx+dx[i]>=0&&yy+dy[i]>=0&&xx+dx[i]<m&&yy+dy[i]<n){vis[xx+dx[i]][yy+dy[i]]=true;ans++;//printf("%d %d\n",xx+dx[i],yy+dy[i]);F(xx+dx[i],yy+dy[i]);}}}int main(){while(scanf("%d%d",&n,&m)&&(n||m)){memset(vis,false,sizeof(vis));for(int i=0;i<m;i++){getchar();for(int j=0;j<n;j++){scanf("%c",&map[i][j]);if(map[i][j]=='@'){x=i,y=j;vis[i][j]=true;}}}ans=0;F(x,y);printf("%d\n",ans+1);}return 0;}