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

---

1. 题意：给定字符表，问有多少个点与@字符相连通。
2. 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。
3. 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。
4. 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; }