POJ1979题解
Red and Black
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 25024 | Accepted: 13511 |
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.
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)
The end of the input is indicated by a line consisting of two zeros.
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)
The end of the input is indicated by a line consisting of two zeros.
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
这道题我用的是dfs算法，不断地找周围的'.'，找到一个后就标记位'#'，开始要记得获得初始的位置。
我的代码：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<cstdlib>
#include<set>
#include<map>
#include<vector>
#include<iomanip>
const int MAX=200;
using namespace std;
char arr[MAX][MAX];
int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
int w,h;
int sum=1;
int dfs(int x,int y)
{
arr[x][y]='#';
for(int k=0;k<4;k++)
{
int dx=x+dir[k][0];
int dy=y+dir[k][1];
if(0<dx&&dx<=h&&0<dy&&dy<=w&&arr[dx][dy]=='.')
{
sum++;
dfs(dx,dy);
}
}
return 0;
}
int main()
{
while(scanf("%d%d",&w,&h)&&w&&h)
{
int beg,end;
for(int i=1;i<=h;++i)
{
for(int j=1;j<=w;++j)
{
cin>>arr[i][j];
if(arr[i][j]=='@')
{
beg=i;
end=j;
}
}
}
dfs(beg,end);
printf("%d\n",sum);
sum=1;
}
return 0;
}
写这个代码中遇见的一些问题：
1、无故地提醒我count是ambiguous的，不知道这是为什么，然后我就把count改成了sum了；
2、记得每次要把sum重新初始化，所以我在最后加入了sum=1这句；
3、开始敲的时候，老是分不清w，h，所以行和列数，我搞错了。
欢迎大家一起交流(*^__^*) 嘻嘻……