HDU-1312-Red and Black

I - Red and Black

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit

Status

Practice

HDU 1312

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

题意，.代表可通行#代表不可通行@代表起始点且可通行，输出一空可以到达多少个.

深度搜索用递归，不过那个输入地图大小的顺序真是坑啊，刚开始都没发现，样例输入居然是倒着输入地图大小

代码

#include<stdio.h>
#include<string.h>
#include<string>
#include<stack>
#include<queue>
#include<math.h>
#include<limits.h>
#include<iostream>
#include<algorithm>
using namespace std;
//深度搜索
char map[22][22];
int dir[4][2]= {{0,1},{0,-1},{1,0},{-1,0}};
int m,n;//地图边界
int starx,stary;//起始点坐标
int num;//记录能到达的数量
void DFS(int starx,int stary)
{
num++;
map[starx][stary]='#';
for(int i=0; i<4; i++)
{
int endx=starx+dir[i][0];
int endy=stary+dir[i][1];
if(endx>=0&&endx<m&&endy>=0&&endy<n&&map[endx][endy]=='.')//坐标合法且可通行
DFS(endx,endy);
}
return;
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
if(m==0&&n==0)
break;
for(int i=0; i<m; i++)
{
for(int j=0; j<n; j++)
{
cin>>map[i][j];
if(map[i][j]=='@')
{
starx=i;
stary=j;
}
}
}
num=0;
DFS(starx,stary);
printf("%d\n",num);
}
}