A - Red and Black POJ 1979

A - Red and Black

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

Submit

Status

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

….#.

…..#

……

……

……

……

……

@…# 此处数据请看原题

.#..#.

11 9

.#………

.#.#######.

.#.#…..#.

.#.#.###.#.

.#.#..@#.#.

.#.#####.#.

.#…….#.

.#########.

………..

11 6

..#..#..#..

..#..#..#..

..#..#..###

..#..#..#@.

..#..#..#..

..#..#..#..

7 7

..#.#..

..#.#..

.

…@…

.

..#.#..

..#.#..

0 0

Sample Output

45

59

6

13

题意：给定字符表，问有多少个点与@字符相连通。

思路：从起始点开始，进行搜索，搜索分四个方向，如果搜到的满足条件，记录值加一，并以此点为根节点继续向四个方向搜索；搜索到不满足的点直接返回。

失误：用广搜的思路没有错，条件主干代码也没有错，主要是一些小的问题检查耗费了时间：数组越界，变量用反了，字符输入了。形式上的错误·还好找，逻辑上的错误就不好找到了。

代码如下:

#include<cstdio>
#include<cstring>

const  int INF=0xfffffff,MAXN=1e2+10;
char  a[MAXN][MAXN];//MAXN不能超过10^4，再大就要用其他的数据结构存储了
bool vis[MAXN][MAXN];
int cnt,xf[5]={0,0,0,-1,1},yf[5]={0,1,-1,0,0};//移动的四个方向
int l,h,sx,sy;

void DFS(int x,int y)//搜索能走多少个
{
if(x>=1&&y>=1&&a[x][y]!='#'&&!vis[x][y]&&x<=h&&y<=l)//满足根节点的条件 边界有四条 不能少写
{                                                  //对于某些情况可初始化成墙
++cnt;//记录走过多少个了
vis[x][y]=true;//以当前节点为根节点
for(int i=1;i<=4;++i)
{
DFS(x+xf[i],y+yf[i]);//将子问题看成与原问题相同的结构
}
}
return ;//递归必须有返回条件
}

int main()
{

while(~scanf("%d%d",&l,&h)&&(h||l))//以后要注意了 即使题里面的输入没有说0怎么办
{                                 //并且测试数据中有0 应该 引起注意
getchar();
memset(vis,false,sizeof(vis));
for(int i=1;i<=h;++i)
{
for(int j=1;j<=l;++j)
{
scanf("%c",&a[i][j]);

if(a[i][j]=='@')//找出初始坐标
{
sx=i;
sy=j;
}
}
getchar();//容易忽略 造成输出的格式都不对
}

cnt=0;
DFS(sx,sy);
printf("%d\n",cnt);
}
return 0;
}