Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25397 Accepted Submission(s): 15306

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

45 59 6 13

Source

Asia 2004, Ehime (Japan), Japan Domestic

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分析：

只能上下左右四个方向走，问你可以走的块最多是多少？#不能走

小技巧：走过的地方字符就变为#

先用dfs写一下，有时间再用bfs写

code：

#include<bits/stdc++.h>
using namespace std;
#define max_v 25
char G[max_v][max_v];
int n,m;
int sx,sy;
int step;
int dir[4][2]={0,1,1,0,0,-1,-1,0};
void dfs(int x,int y)
{int xx,yy;for(int i=0;i<4;i++){xx=x+dir[i][0];yy=y+dir[i][1];if(xx>=0&&xx<n&&yy>=0&&yy<m&&G[xx][yy]!='#'){step++;G[xx][yy]='#';dfs(xx,yy);}}
}
int main()
{while(~scanf("%d %d",&m,&n)){if(n==0&&m==0)break;getchar();for(int i=0;i<n;i++){for(int j=0;j<m;j++){cin>>G[i][j];if(G[i][j]=='@'){sx=i;sy=j;}}}step=1;G[sx][sy]='#';dfs(sx,sy);cout<<step<<endl;}return 0;
}

转载于:https://www.cnblogs.com/yinbiao/p/9350674.html

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