Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11542 Accepted Submission(s): 7185
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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
解题思路
这是一道最基础的深搜，也是朋友教给我的，以前都是不懂。
深搜就是从一个点向上下左右四个方向拓展，然后再以拓展的点为基础向上下左右拓展判断。
所以需要两个数组，bianx[4]和biany[4]来存储x和y的对应变化，一定要注意x、y的变话要对照。
然后还需要判断地点有意义，即在范围之内1<=x<=n&&1<=y<=m
解题代码
#include<cstdio>
#include<cstring>
#include<iostream>//
using namespace std;//C++要加这两个头文件
char map[22][22];
int bx[5]={0,1,0,-1};
int by[5]={1,0,-1,0};
int sum;
int n,m;
bool judge(int a,int b)
{
if(a<1||a>m||b<1||b>n)
return false;
else
{
if(map[a][b]!='#')
return true;
else
return false;
}
}
void dfs(int a,int b)
{
int i;
int nowa,nowb;
sum++;
map[a][b]='#';
for(i=0;i<4;i++)
{//对定点进行上下左右四种变化
nowa=a+bx[i];
nowb=b+by[i];
if(judge(nowa,nowb))
dfs(nowa,nowb);
//直接递归调用就好
else
continue;
}
}
int main()
{
//int n,m;
int i,j,k;
int stax,stay;
while(scanf("%d%d",&n,&m),n+m)
{
memset(map,0,sizeof(map));
for(i=1;i<=m;i++)
for(j=1;j<=n;j++)
{
cin>>map[i][j];
if(map[i][j]=='@')
{
stax=i;
stay=j;
}
}
sum=0;
dfs(stax,stay);
cout<<sum<<endl;
//若输出需要换行，则加上<<endl
}
return 0;
}