1504061021-hd-Red and Black

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.#..........#.#######..#.#.....#..#.#.###.#..#.#..@#.#..#.#####.#..#.......#..#########............11 6..#..#..#....#..#..#....#..#..###..#..#..#@...#..#..#....#..#..#..7 7..#.#....#.#..###.###...@...###.###..#.#....#.#..0 0

 

Sample Output

4559613

 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;}