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.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.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題解：本題還是DFS搜索（上下左右），只是增加一個計數器，計算可以走的地方的個數。 規定地圖中有可通行的位置，也有不可通行的位置，已知起點，求起點的與它相連成一片的部分，在這道題裡輸出相連的位置的數目。 從起點開始，遍歷每一個到達的點的四個方向（不再是八個），到達一個位置就將這個位置的字符變成不可走的'#'，並且計數+1。其實就是計數將可走變成不可走的操作進行了多少次。 這樣可以不用擔心走過了還會重復。 AC 代碼：如果你看過我的上一篇你一定會懂
#include<cstdio>
#include<cstring>
char pic[110][110];
int m,n,total;
int idx[110][110];
void dfs(int r,int c,int id)
{
if(r<0||r>=m||c<0||c>=n)
return;
if(idx[r][c]==666||pic[r][c]!='.')
return;
idx[r][c]=id;
total++;
for(int dr=-1; dr<=1; dr++)
for(int dc=-1; dc<=1; dc++)
if(dr==0||dc==0)
dfs(r+dr,c+dc,id);
}
int main()
{
int i,j;
while(scanf("%d%d",&n,&m)==2&&m&&n)
{
for(i =0; i<m; i++)
scanf("%s",pic[i]);
memset(idx,0,sizeof(idx));
total=0;
for(i=0; i<m; i++)
for(j=0; j<n; j++)
{
if(pic[i][j]=='@')
{
pic[i][j]='.';
dfs(i,j,666);
}
}
printf("%d\n",total);
}
return 0;
}
這個是我在其他博客上看得到的方法，用#填充，可以一試！
#include <iostream>
using namespace std;
char a[25][25];
int n,m,total;
int dr[4] = {0,1,0,-1};//行變化
int dc[4] = {1,0,-1,0};//列變化
//上面的原來一直不會用，知道的話非常方便
bool judge(int x,int y)
{
if(x<1 || x>n || y<1 || y>m)
return 1;
if(a[x][y]=='#')
return 1;
return 0;
}
void dfs(int r,int c)
{
total++;
a[r][c]='#'; //走過一次，“。”變為“#”,避免重復
for(int k=0; k<4; k++)
{
int lr = r + dr[k];
int lc = c + dc[k];
if(judge(lr,lc))
continue;
dfs(lr,lc);
}
}
int main()
{
while(cin>>m>>n&&m&&n)
{
int i,j,x,y;
total=0;
for(i=1; i<=n; i++)
for(j=1; j<=m; j++)
{
cin>>a[i][j];
if(a[i][j]=='@') //這裡必須用變量x，y
x=i,y=j;
}
dfs(x,y);
cout<<total<<endl;
}
return 0;
}
