hdu1312 Red and Black（入门dfs）

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13080    Accepted Submission(s): 8109

Problem 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

题目很简单，注意W和H别搞混了就行，仔细看题

【源代码】

#include<iostream>#include<cstring>using namespace std;int n,m;char map[25][25];bool vis[25][25];int stx,sty,cnt;int dx[4]={0,0,1,-1};int dy[4]={1,-1,0,0};bool judge(int x,int y){//将判断条件抽取if(x<0||x>=n||y<0||y>=m||vis[x][y]||map[x][y]=='#')return false;return true;}void dfs(int x,int y){for(int i=0;i<4;i++){int next_x=x+dx[i];int next_y=y+dy[i];if(!judge(next_x,next_y)) continue;vis[next_x][next_y]=true;cnt++;//cout<<next_x<<" "<<next_y<<endl;dfs(next_x,next_y);}}int main(){while(cin>>m>>n&&(n||m)){for(int i=0;i<n;i++)for(int j=0;j<m;j++){cin>>map[i][j];if(map[i][j]=='@'){stx=i;sty=j;}}cnt=1;memset(vis,0,sizeof(vis));vis[stx][sty]=true;dfs(stx,sty);cout<<cnt<<endl;}return 0;}